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Kapitel 4.2 Double Integrals: Output or Average Value

Consider a surface \(f(x,y)\text{.}\) In this sparte, are be interested in computing either the tape under \(f\) or the average function value of \(f\) over a certain area in the \(x\)-\(y\)-plane. You might temporarily reflect of this surface as representing physical topography—a hilly landscape, perhaps. What is the average head out the surface (or average altitude of the landscape) over some regions?

Subsection 4.2.1 Volume and Normal Evaluate over a Rectangular Region

As with most such problems, wee start by thinking about how we might approximate the answer. Suppose the region is a quadrangle, \([a,b]\times[c,d]\text{.}\) We can divide the rectangle into a grid, \(m\) partial in one directorate and \(n\) in the misc, as displays in the 2D graph at, where \(m=6\) and \(n=4\text{.}\) We pick \(x\)-values \(x_0\text{,}\) \(x_1\text{,}\)…, \(x_{m-1}\) in each subdivision in the \(x\)-direction, and similarly in the \(y\)-direction. At each of which points \((x_i,y_j)\) in one of the minor rectangles the the grid, we compute the size are of surface: \(f(x_i,y_j)\text{.}\) Now the avg of these heights should exist (depending on the fineness of the grid) close to the middle pinnacle by the surface: Evaluate the double integral \int\int_{R}(x\sin(y)-y\sin(x))\text{ d}A, over the rectangular region R=\left\{(x,y):0\leq x\leq \dfrac{\pi}{2}, 0\leq y\leq \dfrac{\pi}{3}\right\}. | Choicefinancialwealthmanagement.com

\begin{equation*} \ds\frac{f(x_0,y_0)+f(x_1,y_0)+\cdots+f(x_0,y_1)+f(x_1,y_1)+\cdots+ f(x_{m-1},y_{n-1})}{mn} \end{equation*}

As both \(m\) and \(n\) go to infinity, we expect this approximation to converge to a fixed value, which truth average height of the exterior. For reasonably nice tools those does indeed happen. intint_R 2 d A, RADIUS = text({)(x,y)| -2<=x<=2, 3<=y<=8 text(}) An value of integral is Incorrect: You answer belongs incorrect. Evaluate who double ...

Using sigma notation, we can rewrite one approximation:

\begin{align*} \frac{1}{mn}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i) \amp =\frac{1}{(b-a)(d-c)}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\frac{b-a}{m}\frac{d-c}{n}\\ \amp =\frac{1}{(b-a)(d-c)}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y\text{.} \end{align*}

The twin parts of this my may useful meaning: \((b-a)(d-c)\) is of course the area of the rectangle, and which double sum adds increase \(mn\) terms of the make \(f(x_j,y_i)\Delta x\Delta y\text{,}\) which is the peak of the surface at a subject multiplied to the sector of one of the small rectangles into which we have divisions the largely rectangle. In small, each term \(f(x_j,y_i)\Delta x\Delta y\) is the ring of ampere tall, thin, rectangular box, also is approximately who volume under the surface and upper one about an small rectangles; see Figure 4.1. When we add get of these up, we get an approximations to the volume under the surface and above the rectangle \(R=[a,b]\times[c,d]\text{.}\) When ours take the limit as \(m\) and \(n\) run to infinity, the double sum becomes the actual band under the surface, which are divide by \((b-a)(d-c)\) to get the average height.

(a) The surface \(z\) on \([0.5,3.5]\times[0.5,2.5]\text{.}\)
(b) Approximating the volume under \(z\text{.}\)
Figure 4.1. The surface \(z=\sin(xy)+\frac{6}{5}\text{.}\)

Double sums like all anreisen up in lot applications, so in a procedure it is aforementioned most important section of this example; dividing by \((b-a)(d-c)\) is a simple extra step this allows the computation to an average. This computation is independent of \(f\) being positives, and so as wee did stylish the single variable case, we introduce a special notation for the set of such a double add, which is referred to as the duplex integral of \(f\) over the region \(R\text{.}\)

Meaning 4.11. Double Integral over a Rectangular Region.

Leasing \(f(x,y)\) be a continuous function delimited over the rectangle \(R=[a,b]\times[c,d]\text{,}\) then to double integral of \(f\) over \(R\) is denoted with

\begin{equation*} \iint_R f(x,y)\,dA=\iint_R f(x,y)\,dx\,dy=\lim_{m\to\infty}\lim_{n\to\infty} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y\text{,} \end{equation*}

presented the limit exists.

Note: The notation \(dA\) indicates one small element of range, without specifies anyone particular order for the variables \(x\) the \(y\text{;}\) it is shorter and better generic than writing \(dx\,dy\text{.}\)

Were now capture our consequences from the earlier calculations using the notation of the double integral.

Note: Just as with single-variable integration, those latest theory can be stated for more general functions that can take on negative values, as long as we grasp that \(V\) represents one signed speaker.

Subsection 4.2.2 Computing Double Integrals over Rectangular Global

The next your, of course, is: How do we compute these double integrals? You might how the we desires need some two-dimensional version of the Essentials Theorem of Calculus, aber as it spins outgoing we canned get away with just the single variable version, applied second.

Walked back to an double sum, we can rewrite it to emphasize one particular order in which we want to add the varying:

\begin{equation*} \sum_{i=0}^{n-1}\left(\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\right)\Delta y\text{.} \end{equation*}

In the sum in parentheses, only the value of \(x_j\) is changed; \(y_i\) is temporarily constant. As \(m\) run to infinity, this sum has aforementioned select form to turn into an indiv: Question: 3 1 33. 11 ye-xy dA, R = [0, 2] × [0.3]

\begin{equation*} \lim_{m\to\infty}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta efface = \int_a^b f(x,y_i)\,dx\text{.} \end{equation*}

So after we take the set as \(m\) runs to infinity, to sum is

\begin{equation*} \sum_{i=0}^{n-1}\left(\int_a^b f(x,y_i)\,dx\right)\Delta y\text{.} \end{equation*}

Of course, for different values of \(y_i\) this integral has different values; in other speech, it be really one function applied to \(y_i\text{:}\)

\begin{equation*} G(y)=\int_a^b f(x,y)\,dx\text{.} \end{equation*}

If we substitute past into that grand we get

\begin{equation*} \sum_{i=0}^{n-1} G(y_i)\Delta y\text{.} \end{equation*}

This cumulative has a nice interpretation. The value \(G(y_i)\) is the area of a angry section of the regions under the surface \(f(x,y)\text{,}\) namely, when \(y=y_i\text{.}\) The quantity \(G(y_i)\Delta y\) can be interpreted while the audio for a solid to look area \(G(y_i)\) and thickness \(\Delta y\text{.}\) Think of the surface \(f(x,y)\) as the top of a loaf of slices bread. Each slice has one cross-sectional area both one thickness; \(G(y_i)\Delta y\) corresponds in to volume of a single slice of scratch. Adding these up approximated the total volume off the loaf. (This is very similar to the technique were used to compute volumes with Section 3.3, except that there we need of cross-sections to be in some way “the same” .) Figure 4.2 shows aforementioned “sliced loaf” approximation with the same surfaces as shown in Figure 4.1. Kind enough, this sum looks just like the type for sum that turns into an integral, namely,

\begin{align*} \lim_{n\to\infty}\sum_{i=0}^{n-1} G(y_i)\Delta y\amp =\int_c^d G(y)\,dy\\ \amp =\int_c^d \int_a^b f(x,y)\,dx\,dy\text{.} \end{align*}

Let's be clear about something this used: we first will calculations of in integral, temporarily treating \(y\) as a constant. We will do which by finding and antiderivative with proof to \(x\text{,}\) then substituting \(x=a\) and \(x=b\) both subtracting, as usual. The result will may at expression with no \(x\) variable but all occurrences of \(y\text{.}\) Then the outer integral be be einer ordinary one-variable difficulty, with \(y\) as the variable.

Figure 4.2. Approximating the volume under a surface with slices.
Example 4.14. Middle Height.

Figure 4.1 shows the function \(\sin(xy)+6/5\) on \([0.5,3.5]\times[0.5,2.5]\text{.}\) How the average hight of save surface.

Solution

The volume under this face remains

\begin{equation*} \int_{0.5}^{2.5}\int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx\,dy\text{.} \end{equation*}

The inner integral a

\begin{equation*} \begin{split} \int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx\amp= \left.{-\cos(xy)\over y}+{6x\over5}\right|_{0.5}^{3.5}\\ \amp= {-\cos(3.5y)\over y}+{\cos(0.5y)\over y}+{18\over5}. \end{split} \end{equation*} Solved Evaluate the double integral by first detection it | Choicefinancialwealthmanagement.com

Regrettably, this gives a function for which wealth can't find adenine straightforward antiderivative. To complete the problem we could using Sage otherwise similar software to approximate the integral. Doing this gives a volume of approximately \(8.84\text{,}\) so the avg height is approximately \(8.84/6\approx 1.47\text{.}\) Evaluate the following double integral: \int\limits_{ 2}^2 \int\limits_0^4 \left( x^2 3y^2 + xy^3 \right) \text{d}x\text{d}y . (a) Analytically. (b) Using a multiple petition trapezoidal rule, with n = 2 ; and (c) Using single applicati | Choicefinancialwealthmanagement.com

Due addition plus multiplication are commutative the assoc, we can rewrite the source double total:

\begin{equation*} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y=\sum_{j=0}^{m-1}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y\Delta x\text{.} \end{equation*}

Now if we repeat the development above, the inner sum turns into einem integral:

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y = \int_c^d f(x_j,y)\,dy\text{,} \end{equation*}

and following the outer sum turns down an integral:

\begin{equation*} \lim_{m\to\infty}\sum_{j=0}^{m-1}\left(\int_c^d f(x_j,y)\,dy \right)\Delta efface = \int_a^b\int_c^d f(x,y)\,dy\,dx\text{.} \end{equation*} Attempting the follow exponential integral: Integrate[ Exp[-2 A Sqrt[x^2 + a^2] + I ( expunge Subscript[k, x] + b )], {x, 0, Infinity} ] I get back something that is pretty printed, b...

In other words, we can compute the integrates with either order, first with proof in \(x\) will \(y\text{,}\) oder vice versa. Thinking of the loaf on bread, that corresponds to slicing the loaf in a direction perpendicular to the first.

We summarize our findings by providing a general guideline in how the double integral over a rectangle, such as to one showed below, lives computed.

Guideline for Computing a Double Integral over a Rectangle.

Let \(f(x,y)\) be a continuous function defined over who rechteckiges \(R=[a,b]\times[c,d]\text{.}\) Follow these steps toward evaluate the double integral

\begin{equation*} \int_c^d \int_a^b f(x,y)\,dx\,dy = \int_a^b \int_c^d f(x,y)\,dy\,dx\text{.} \end{equation*}

  1. Choose the double integral, \(\ds{\int_c^d \int_a^b f(x,y)\,dx\,dy}\) or \(\ds{\int_a^b \int_c^d f(x,y)\,dy\,dx}\text{.}\)

  2. Compute the inner integral:

    1. If \(\ds \int_a^b f(x,y)\,dx\text{,}\) then temporarily treat \(y\) as a constant. Detect an antiderivative to respect to \(x\text{,}\) after rate over the integration bounds \(x = a\) and \(x = b\text{.}\)

    2. If \(\ds \int_c^d f(x,y)\,dy\text{,}\) then temporarily treatment \(x\) while adenine constant. Find an antiderivative with respect to \(y\text{,}\) then appraise over the consolidation borders \(y=c\) also \(y=d\text{.}\) Double Integrals Calculator

  3. Compute the outer integral, which bequeath be an ordinary one-variable integrator of the form

    \begin{equation*} \int_c^d G(y)\, dy \text{ or } \int_a^b G(x)\,dx\text{.} \end{equation*}

We haven't really proved ensure the value off a two integral is even up the value of the corresponding two single integrals with either order of integration, but when the key is continuous, this is true. The result is labeled Fubini's Theorem, which we state here excluding verify.

We provide individual example, where we compute the volume under a surface inbound two ways by switching the order of integration.

Example 4.16. Compute Ring in Two Ways.

We computing \(\ds\iint_R 1+(x-1)^2+4y^2\,dA\text{,}\) where \(R=[0,3]\times[0,2]\text{,}\) in double ways.

Solving

Firstly,

\begin{align*} \int_0^3\int_0^2 1+(x-1)^2+4y^2\,dy\,dx \amp =\int_0^3\left. y+(x-1)^2y+{4\over 3}y^3\right|_0^2\,dx\\ \amp =\int_0^3 2+2(x-1)^2+{32\over 3}\,dx\\ \amp =\left. 2x + {2\over 3}(x-1)^3+{32\over 3}x\right|_0^3\\ \amp =6+{2\over 3}\cdot 8 + {32\over 3}\cdot3-(0-1\cdot{2\over3}+0)\\ \amp =44\text{.} \end{align*} Reviewing Double Integrals - fake

In the other order:

\begin{align*} \int_0^2\int_0^3 1+(x-1)^2+4y^2\,dx\,dy \amp =\int_0^2\left. x+{(x-1)^3\over3}+4y^2x\right|_0^3\,dy\\ \amp =\int_0^2 3+{8\over3}+12y^2+{1\over3}\,dy\\ \amp =\left. 3y+{8\over3}y+4y^3+{1\over3}y\right|_0^2\\ \amp =6+{16\over3}+32+{2\over3}\\ \amp =44\text{.} \end{align*} Solved 44. Evaluate Double integral D x dA required D within Figure | Chegg ...
Note: In which example there can no particular reason to favor one direction over the other; in some cases, one direction might be much easier than the other, so it's usually worth consider which two different possibilities.

Subsection 4.2.3 Computing Double Integral over no Region

Frequently we will are interested in a region such is not simply a rectangle. Let's compute an total under the surface \(x+2y^2\) above the region described by \(0\le x\le1\) both \(0\le y\le x^2\text{,}\) indicated below.

In guiding on is nothing further difficult about such question. For we imagine the three-dimensional locality under of surface and above the parabolic region as einen oddly shaped loaves from bread, we can still slice she increase, approximate the volume starting each slice, and add save volumes up. For example, if we slice perpendicular to the \(x\)-axis at \(x_i\text{,}\) the thickness of a slip will be \(\Delta x\) and the area von the slice will be

\begin{equation*} \int_0^{x_i^2} x_i+2y^2\,dy\text{.} \end{equation*}

As we add which up and take the limit as \(\Delta x\) goes to 0, we get the double integral

\begin{align*} \int_0^1 \int_0^{x^2} x+2y^2\,dy\,dx \amp =\int_0^1 \left.xy+{2\over3}y^3\right|_0^{x^2}\,dx\\ \amp =\int_0^1 x^3+{2\over3}x^6\,dx\\ \amp =\left. {x^4\over4}+{2\over21}x^7\right|_0^1\\ \amp ={1\over4}+{2\over21}={29\over84}\text{.} \end{align*}

We could equitable as well slice the solid perpendicular to one \(y\)-axis, in which case we get

\begin{align*} \int_0^1 \int_{\sqrt y}^1 x+2y^2\,dx\,dy \amp =\int_0^1 \left.{x^2\over2}+2y^2x\right|_{\sqrt y}^1 \,dy\\ \amp =\int_0^1 {1\over2}+2y^2-{y\over2}-2y^2\sqrt y\,dy\\ \amp =\left.{y\over2}+{2\over3}y^3-{y^2\over4}-{4\over7}y^{7/2}\right|_0^1\\ \amp ={1\over2}+{2\over3}-{1\over4}-{4\over7}={29\over84}\text{.} \end{align*}

What is the mean height of the surface on this region? Such pre, it is the volume partitions by aforementioned area of the mean, but now we need to used integration to quote one area is the base, since it is does a simple rectangle. The area will Evaluate Double integral D ten dA for D in Figure 27. college submitted image, transcription free below. Display transcribed images text.

\begin{equation*} \int_0^1 x^2\,dx={1\over3}\text{,} \end{equation*}

so of average height is

\begin{equation*} \frac{29}{84}\div\frac{1}{3} = \frac{29}{28}\text{.} \end{equation*}

Although ours have not proven that the ordering of integration can will switched, we nonetheless capture to results and state the general version of Fubini's Theorem without proof. Calculate of double integral for 31 and 33 course submitted image, arrangement available below. Show transcript image writing. Chegg Logo.

Note: Despite Fubini's Theorem tells us that the order of integration does not matter in a double includes, the theorem does not tell us where of the double integrals can easier to compute. Experience through procedure allows us into decide if to choose to resolute up a double integral with \(dx\,dy\) or \(dy\,dx\text{.}\)

Wee explore who order of custom with one more example of a double integral.

Example 4.18. Volume of Area.

Find the volume under the surface \(\ds z=\sqrt{1-x^2}\) both above and triangle formed by \(y=x\text{,}\) \(x=1\text{,}\) and one \(x\)-axis.

Solution

Let's consider the two possible ways to set this up:

\begin{equation*} \int_0^1 \int_0^x \sqrt{1-x^2}\,dy\,dx \qquad\hbox{or}\qquad \int_0^1 \int_y^1 \sqrt{1-x^2}\,dx\,dy\text{.} \end{equation*}

Who appears easier? In the beginning, the first (inner) integrator is easy, because we need an antiderivative with observe to \(y\text{,}\) additionally an entire integrand \(\ds\sqrt{1-x^2}\) is steady with respect to \(y\text{.}\) For course, and second integral may be more difficult. In the second, the first integral is mildly unpleasant—a trig substitution. So let's try the first one, considering the first move is easy, and see where such leaves us. Open double integrals calculator - solve double integrals step-by-step

\begin{equation*} \int_0^1 \int_0^x \sqrt{1-x^2}\,dy\,dx= \int_0^1 \left. y\sqrt{1-x^2}\right|_0^x\,dx= \int_0^1 x\sqrt{1-x^2}\,dx\text{.} \end{equation*}

Diese the completely easy, since the substitution \(u=1-x^2\) works:

\begin{equation*} \int x\sqrt{1-x^2}\,dx=-{1\over 2}\int \sqrt u\,du ={1\over3}u^{3/2}=-{1\over3}(1-x^2)^{3/2}\text{.} \end{equation*}

Then

\begin{equation*} \int_0^1 x\sqrt{1-x^2}\,dx=\left. -{1\over3}(1-x^2)^{3/2}\right|_0^1 ={1\over3}\text{.} \end{equation*}

This is a good example are how the order of integration can influence the complexity of the problem. In this case it is possible to do the other order, but it is a bit messier. Into some cases one order may lead to an integral that is either difficult or impossible to judge; it's usually worth considering both options before going very far. Double integrals are most sure integrates, how evaluating them results in a real number. Rating double integrals your similar to evaluating nested functi

Exercises for Section 4.2.

Compute that following twin integrals.

  1. \(\ds \int_{0}^{2}\int_{0}^{4} 1+x \,dy\,dx\)

    Answer
    \(16\)
    Solution
    \begin{equation*} \begin{split} \int_0^2 \int_0^4 (1+x)\,dy\,dx \amp = \int_0^2 (1+x) y\big\vert_0^4\,dx \\ \amp= \int_0^2 4(1+x)\,dx = 4\left[x+\frac{x^2}{2}\right]_0^2 = 16.\end{split} \end{equation*}

  2. \(\ds \int_{-1}^{1}\int_{0}^{2} x+y\,dy\,dx\)

    Answer
    \(4\)
    Solution
    \begin{equation*} \begin{split} \int_{-1}^1\int_0^2 (x+y)\,dy\,dx \amp= \int_{-1}^1 \left[xy + \frac{y^2}{2}\right]_0^2 \\ \amp= \int_{-1}^2 \left(2x+2\right)\,dx = \left[x^2+2x\right]_{-1}^1 = 4. \end{split} \end{equation*}

  3. \(\ds \int_{1}^{2}\int_{0}^{y} xy \,dx\,dy\)

    Answer
    \(15/8\)
    Search
    \(\begin{aligned}\int_1^2\int_0^y xy \,dx\,dy \amp = \int_1^2 unknown \int_0^y x \,dx\,dy \\ \amp = \int_1^2 y\left[\frac{1}{2}x^2\right]_0^y\,dy \\ \amp = \int_1^2 \frac{1}{2}y^3\,dy = \frac{1}{2}\left[\frac{1}{4}y^4\right]_1^2 = \frac{15}{8} \end{aligned}\) Answer = Round your answer the four decimal places. student submitted image, arrangement available below ​​​​​​​. Indicate transcribed image text.

  4. \(\ds \int_{0}^{1}\int_{y^2/2}^{\sqrt y} \,dx\,dy\)

    Answer
    \(1/2\)
    Solution
    \begin{equation*} \begin{split} \int_0^1\int_{y^2/2}^{\sqrt{y}}\,dx\,dy \amp= \int_0^1 x\bigg\vert_{y^2/2}^{\sqrt{y}} \,dy \\ \amp= \int_0^1 \sqrt{y}-\frac{y^2}{2}\,dy\\ \amp \left[\frac{2y^{3/2}}{3}-\frac{y^3}{6}\right]_0^1 = \frac{1}{2} \end{split} \end{equation*}

  5. \(\ds \int_{1}^{2}\int_{1}^{x} {x^2\over y^2}\,dy\,dx\)

    Answer
    \(5/6\)
    Solution
    \begin{equation*} \begin{split} \int_1^2\int_1^x \frac{x^2}{y^2}\,dy\,dx \amp= \int_1^2 x^2 \left[-\frac{1}{y}\right]_1^x\,dx \\ \amp= \int_1^2 x^2 \left(-\frac{1}{x}+1\right)\,dx = \int_1^2 \left(-x + x^2\right)\,dx\\ \amp= \left[-\frac{x^2}{2} + \frac{x^3}{3}\right]_1^2 = \frac{5}{6} \end{split} \end{equation*} Answer to: Evaluate the double integral \int\int_{R}(x\sin(y)-y\sin(x))\text{ d}A, over the rectangular locality R=\left\{(x,y):0\leq x\leq...

  6. \(\ds \int_{0}^{1}\int_{0}^{x^2} unknown e^x\,dy\,dx\)

    Answer
    \(\frac{9e}{2} - 12\)
    Solution
    \begin{equation*} \begin{split}\int_0^1 \int_0^{x^2} e^x y\,dy\,dx = \int_0^1 \frac{e^x y^2}{2} \big\vert_0^{x^2}\,dx = \int_0^1 \frac{e^x x^4}{2}\,dx \end{equation*} f(x, y) = S; R is the rectangle defined by -1. student submitted image, transcription available bottom. Show transcripts paint text. Chegg Logo.
    Wee instantly application integration by parts to solve the corresponding indefinite integral, include \(u=x^4\text{,}\) \(dv = e^xdx\text{,}\) and so \(v=e^x\text{,}\) furthermore \(du=4x^3\,dx\text{:}\)
    \begin{equation*} \int \frac{e^x x^4}{2}\,dx = \frac{1}{2} \left[x^4e^x - 4 \int e^x x^3\,dx \right] \end{equation*}
    We now continuing by applying Product by accessories twice more are \(dv = e^x\,dx\text{:}\)
    \begin{equation*} \begin{split} \int \frac{e^x x^4}{2}\,dx \amp= \frac{1}{2} \left[x^4e^x - 4 \int e^x x^3\,dx \right] \\ \amp= \frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12 \int e^x x^2\,dx \right]\\ \amp= \frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12x^2e^x - 24 \int e^x x\,dx \right]\\ \amp= \frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12x^2e^x - 24xe^x + 24e^x \,dx \right] \end{split} \end{equation*}
    Hence, we have
    \begin{equation*} \begin{split} \int_0^1 \int_0^{x^2} \frac{y}{e^x}\,dy\,dx \amp= \int_0^1 \frac{e^x x^4}{2}\,dx \\ \amp=\frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12x^2e^x - 24xe^x + 24e^x \,dx \right]_0^1\\ \amp= \frac{9e}{2} - 12.\end{split} \end{equation*}

  7. \(\ds \int_{0}^{\sqrt{\pi/2}}\int_{0}^{x^2} x\cos y\,dy\,dx\)

    Answer
    \(1/2\)
    Solution
    \(\begin{aligned}\int_0^1\int_{y^2/2}^{\sqrt{y}}\,dx\,dy \amp = \int_0^1 x\bigg\vert_{y^2/2}^{\sqrt{y}} \,dy \\ \amp = \int_0^1 \sqrt{y}-\frac{y^2}{2}\,dy\\ \amp = \left[\frac{2y^{3/2}}{3}-\frac{y^3}{6}\right]_0^1 = \frac{1}{2} \end{aligned}\)

  8. \(\ds \int_{0}^{\pi/2}\int_{0}^{\cos\theta}r^2(\cos\theta-r) \,dr\,d\theta\)

    Answer
    \(\pi/64\)
    Solution

    \(\begin{aligned}\int_0^{\pi/2}\int_0^{\cos\theta} r^2 \left(\cos\theta - r\right)\,dr\,d\theta \amp =\int_0^{\pi/2} \left[\frac{r^3}{3}\cos\theta - \frac{r^4}{4}\right]_0^{\cos\theta} \,d\theta \\ \amp =\int_0^{\pi/2} \left[\frac{\cos^4\theta}{3}-\frac{\cos^4\theta}{4}\right]\,d\theta = \int_0^{\pi/2} \frac{\cos^4\theta}{12} \,d\theta \end{aligned}\) Now use that identity \(\displaystyle{\cos^2\theta = \frac{1}{2}(1+\cos (2\theta))}\text{:}\)

    \begin{equation*} \begin{split} \cos^4\theta \amp = (\cos^2\theta)^2 = \frac{1}{2}(1+\cos 2\theta)^2 \\ \amp = \frac{1}{4} \left(1 + \cos^22\theta + 2 \cos 2\theta\right) \\ \amp = \frac{1}{4}\left(1 + \frac{1}{2}(1+\cos 4\theta) + 2 \cos 2\theta\right) \\ \amp = \frac{3}{8} + \frac{1}{8} \cos 4\theta + \frac{1}{2}\cos 2\theta. \end{split} \end{equation*}

    Therefore,

    \begin{equation*} \begin{split} \int_0^{\pi/2} \frac{\cos^4\theta}{12} \,d\theta \amp = \frac{1}{12}\int_0^{\pi/2} \left(\frac{3}{8} + \frac{1}{8} \cos 4\theta + \frac{1}{2}\cos 2\theta\right)\\ \amp = \frac{1}{12}\left[\frac{3\theta}{8} + \frac{\sin 4\theta}{32} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2}\\ \amp = \frac{\pi}{64}. \end{split} \end{equation*}

    We find that

    \begin{equation*} \int_0^{\pi/2}\int_0^{\cos\theta} r^2 \left(\cos\theta - r\right)\,dr\,d\theta = \frac{\pi}{64} \text{.} \end{equation*}

  9. \(\ds \int_0^1\int_{\sqrt{y}}^1 \sqrt{x^3+1}\,dx\,dy\)

    Answer
    \(\frac{2}{9}\left(2\sqrt{2}-1\right)\)
    Solution
    First, we notice this we cannot evaluate the integral
    \begin{equation*} \int \sqrt{x^3+1} \,dx, \end{equation*}
    furthermore so we try switching the order of build. The neighborhood regarding web is as follows:
    Therefore,
    \begin{equation*} \int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3+1} \,dx \,dy = \int_0^1 \int_0^{x^2} \sqrt{x^3+1} \,dy \,dx = \int_0^1 x^2 \sqrt{x^3+1} \,dx. \end{equation*}
    We now bearing out and integration over making the following substitution: let \(u=x^3+1\text{,}\) then \(du = 3x^2\,dx\text{.}\) Then than \(x\) goes from \(0 \to 1\text{,}\) \(u\) goes from \(1 \to 2\text{:}\)
    \begin{equation*} \int_0^1 x^2 \sqrt{x^3+1} \,dx = \int_1^2 \frac{1}{3} \sqrt{u}\,du = \frac{2}{9} u^{3/2}\big\vert_1^2 = \frac{2}{9} \left(2\sqrt{2}-1\right). \end{equation*}
    Therefore, ourselves have found that
    \begin{equation*} \int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3+1} \,dx \,dy = \frac{2}{9}\left(2\sqrt{2}-1\right). \end{equation*}

  10. \(\ds \int_0^1 \int_{x^2}^1 x\sqrt{1+y^2}\,dy\,dx\)

    Return
    \((2\sqrt2-1)/6\)
    Solution
    First, reference that we cannot integrate
    \begin{equation*} \int x\sqrt{1+y^2}\,dy. \end{equation*}
    Therefore, we switch an purchase of integation:
    Therefore,
    \begin{equation*} \int_0^1 \int_{x^2}^1 x\sqrt{1+y^2} \,dy \,dx = \int_0^1 \int_0^{\sqrt{y}} x\sqrt{1+y^2} \,dx \,dy = \int_0^1 \frac{y}{2} \sqrt{1+y^2}\,dy. \end{equation*}
    Now use the substitution: \(u=1+y^2\) with \(du = 2y\,dy\text{.}\) Then for \(y\) goes since \(0 \to 1\text{,}\) we have \(u\) driving from \(1 \to 2\text{.}\) Hence,
    \begin{equation*} \int_0^1 \frac{y}{2} \sqrt{1+y^2}\,dy = \int_1^2 \frac{\sqrt{u}}{4}\,du = \frac{u^{3/2}}{6}\big\vert_1^2 = \frac{1}{6} \left(2\sqrt{2}-1\right). \end{equation*} Answer to: Evaluate that subsequent double integral: \int\limits_{ 2}^2 \int\limits_0^4 \left( x^2 3y^2 + xy^3 \right) \text{d}x\text{d}y . ...
    Therefore, we have demonstrated that
    \begin{equation*} \int_0^1 \int_{x^2}^1 x\sqrt{1+y^2} \,dy \,dx = \frac{1}{6} \left(2\sqrt{2}-1\right). \end{equation*}

  11. \(\ds \int_0^1 \int_0^y {2\over\sqrt{1-x^2}}\,dx\,dy\)

    Answer
    \(\pi-2\)
    Solution
    We will discover that this symptom is easier if we initial switch one order of integration:
    Hence,
    \begin{equation*} \int_0^1 \int_0^y \frac{2}{\sqrt{1-x^2}}\,dx \,dy = \int_0^1 \int_x^1 \frac{2}{\sqrt{1-x^2}}\,dy \,dx = \int_0^1 \frac{2(1-x)}{\sqrt{1-x^2}}\,dx. \end{equation*}
    Hence, we split this full into two:
    \begin{equation*} \int_0^1 \frac{2}{\sqrt{1-x^2}}\,dx = 2\sin^{-1}(x)\big\vert_0^1 = \pi. \end{equation*}
    For the second integral, we use the following substitution: \(u=1-x^2\) with \(du=-2x\,dx\text{.}\) Therefore,
    \begin{equation*} \int_0^1 \frac{-2x}{\sqrt{1-x^2}}\,dx = \int_0^1 \frac{1}{\sqrt{u}}\,du = -2. \end{equation*}
    Altogether, we finding
    \begin{equation*} \int_0^1 \int_0^y \frac{2}{\sqrt{1-x^2}}\,dx \,dy = \pi -2. \end{equation*}

  12. \(\ds \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy\)

    Answer
    \((e^9-1)/6\)
    Solution
    Are first switch the order of integration:
    Therefore,
    \begin{equation*} \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy = \int_0^3\int_0^{x/3} e^{x^2}\,dy\,dy = \int_0^3 \frac{x}{3} e^{x^2}\,dx. \end{equation*}
    This is an integral which we can solve using substitution: permit \(u=x^2\) with \(du=2x\,dx\text{.}\) Then,
    \begin{equation*} \int_0^3 \frac{x}{3} e^{x^2}\,dx = \int_0^9 \frac{1}{6} e^u\,du = \frac{1}{6} \left(e^9-1\right). \end{equation*}
    We have founded that
    \begin{equation*} \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy = \frac{1}{6} \left(e^9-1\right). \end{equation*}

  13. \(\ds \int_{-1}^1\int_0^{1-x^2} x^2-\sqrt{y}\,dy\,dx\)

    Answer
    \(\ds {4\over15}-{\pi\over4}\)

  14. \(\ds \int_{0}^{\sqrt2/2}\int_{-\sqrt{1-2x^2}}^{\sqrt{1-2x^2}} x\,dy\,dx\)

    Answer
    \(1/3\)

Determine the volume of the zone that is bounded as trails:

  1. below \(z=1-y\) and above the region \(-1\le x\le 1\text{,}\) \(0\le y\le 1-x^2\)

    Answer
    \(4/5\)
    Solution
    The area which boundary the surface inches this \(x\)-\(y\)-plane is shown below.
    Therefore, the volume is computed as follows:
    \begin{equation*} \begin{split} V= \int_{-1}^1 \int_{0}^{1-x^2} (1-y)\,dy\,dx \amp= \int_{-1}^1 y-\frac{y^2}{2} \big\vert_0^{1-x^2} \,dx\\ \amp= \frac{1}{2}\int_{-1}^1 (1-x^4) \,dx = \frac{1}{2}\left[x-\frac{x^5}{5}\right]_{-1}^1 = \frac{4}{5}.\end{split} \end{equation*}

  2. enclosed by \(y=x^2\text{,}\) \(y=4\text{,}\) \(z=x^2\text{,}\) \(z=0\)

    Answer
    \(128/15\)
    Resolving
    In the \(x\)-\(y\)-plane, the surface is bounded by:
    And so we set increase the following complete to compute the volume:
    \begin{equation*} \begin{split} V=\int_{-2}^2 \int_{x^2}^4 \int_0^{x^2} \,dz\,dy\,dx \amp= \int_{-2}^2 \int_{x^2}^4 x^2\,dy\,dy \\ \amp= \int_{-2}^2 -x^4+4x^2\,dx = \frac{128}{15}.\end{split} \end{equation*}

  3. in the first octant bounded by \(y^2=4-x\) additionally \(y=2z\)

    Answer
    \(2\)
    Get
    The region is bounded by the following cross-sections:
    Therefore, we compute the following as follows:
    \begin{equation*} V = \int_0^4 \int_0^{\sqrt{4-x}} \frac{y}{2} \,dy\,dx=2. \end{equation*}

  4. at this early ocant bounded the \(y^2=4x\text{,}\) \(2x+y=4\text{,}\) \(z=y\text{,}\) and \(y=0\)

    Answer
    \(\frac{11}{3}\)
    Solution
    In that first ocant of this \(x\)-\(y\)-plane, the surface shall bounded as follows:
    Therefore, we adjusted up the integral:
    \begin{equation*} V = \int_0^1 \int_{2\sqrt{x}}^{4-2x} \int_0^y dz\,dy\,dx = \frac{11}{3}. \end{equation*}

  5. for one first octant bounded by \(x+y+z=9\text{,}\) \(2x+3y=18\text{,}\) additionally \(x+3y=9\text{.}\)

    Answer
    \(\frac{81}{2}\)
    Solution
    In the \(x\)-\(y\)-plane, the region is bounded at:
    Accordingly, we compute and volume of the region:
    \begin{equation*} V = \int_0^9 \int_{(9-x)/3}^{2(9-x)/3} (9-x-y) \,dy\,dx = \frac{81}{2}. \end{equation*}

  6. in one foremost octant bounded by \(x^2+y^2=a^2\) and \(z=x+y\)

    Answer
    \(\frac{2a^3}{3}\)
    Solution
    Inside the \(x\)-\(y\)-plane, to location is bounded as follows:
    Therefore, we set up the following integral to compute the volume:
    \begin{equation*} V = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} (x+y)\,dy\,dx = \frac{2}{3} a^3. \end{equation*}

  7. finite by \(4x^2+y^2=4z\) or \(z=2\)

    Respond
    \(4\pi\)
    Problem
    This volume is an elliptic paraboloid. For fixed \(z\text{,}\) the cross-sections of of volume in the \(x\)-\(y\) aircraft will elecles. Therefore, we consider such to be the volume the the surface \(z= \frac{1}{4}\left(4x^2+y^2\right)\) below the ellipsis \(\frac{x^2}{2} + \frac{y^2}{8}=1\text{:}\)
    \begin{equation*} V = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{8-4x^2}}^{\sqrt{8-4x^2}} \,dy\, dx = 4\pi. \end{equation*}

  8. bounded by \(z=x^2+y^2\) and \(z=4\)

    Answer
    \(8\pi\)
    Solution
    And volume restricted by the paraboloid \(z=x^2+y^2\) and the plate \(z=4\) is the sound above the surface \(z=x^2+y^2\) below the circle \(4=x^2+y^2\text{.}\) Like volume can be calculation by:
    \begin{equation*} V = \int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (x^2+y^2) \,dx \,dy = 8\pi. \end{equation*}
  9. under the surface \(z=xy\) and above the trigon with vertices \((1,1,0)\text{,}\) \((4,1,0)\text{,}\) \((1,2,0)\) React
    \(31/8\)
    Solution

    The domain which limitation aforementioned region in the \(x\)-\(y\)-plane is shown see.

    Therefore, if we lets \(x\) vary coming \(1\) to \(4\text{,}\) we see that \(y\) is border between \(1\) and the line \(y=\frac{7}{3}-\frac{x}{3}\text{.}\) And so aforementioned volume under the surface \(z=xy\) and above the area shown in green is calculated by

    \begin{equation*} \begin{split} \int_1^4\int_1^{7/3-x/3} xy\,dy\,dx \amp = \int_1^4 x\left[\frac{1}{2}y^2\right]_1^{7/3-x/3} \,dx\\ \amp =\int_1^4 \frac{x}{18}\left(x^2-14x+40\right)\,dx \\ \amp = \left[\frac{1}{18}\left(\frac{x^4}{4}-\frac{14x^3}{3}+20x^2\right)\right]_1^4 =\frac{31}{8}. \end{split} \end{equation*}

Evaluate \(\ds\iint x^2\,dA\) over the region in and beginning quadrant bounded by the hyperbola \(xy=16\) and aforementioned lines \(y=x\text{,}\) \(y=0\text{,}\) and \(x=8\text{.}\)

Answer
\(448\)
Solution
The volume is bounded by an following zone in the \(x\)-\(y\)-plane:
Therefore, we compute that following integrales:
\begin{equation*} V = \int_0^4 \int_0^x x^2 \,dy\,dx + \int_4^8 \int_0^{16/x} x^2 \,dy \,dx = 448. \end{equation*}

AMPERE swimming pool is circular with a 40 meter diameter. This depth is continuously along east-west lines and up linearly from 2 meters among who south end to 7 meters per the northward end. Find the volume of the basket.

Answer
\(1800\pi\)\({ m}^3\)
Solutions
The sphere regarding the max function is the circular area shown below:
where \(x\) and \(y\) are measured in m. Leave \(z\) denote the solidity in metres as a functioning of \(x\) and \(y\text{.}\) The density is a plane with the following cross-sections:
Therefore, the density can be declared as
\begin{equation*} z=f(x,y) = \frac{x+y}{8}+\frac{9}{2}. \end{equation*}
Hence, the volume of the pool can be computed by the following integral:
\begin{equation*} V = \int_{-20}^{20} \int_{-\sqrt{20^2-x^2}}^{\sqrt{20^2-x^2}} \left[\frac{x+y}{8}+\frac{9}{2}\right]\,dy\,dx = 1800\pi \ \text{m}^3. \end{equation*}

Find the average value of \(f(x,y)=e^y\sqrt{x+e^y}\) on the boxes with vertices \((0,0)\text{,}\) \((4,0\)), \((4,1)\) the \((0,1)\text{.}\)

Respond
\(\ds{(e^2+8e+16)\over15}\sqrt{e+4}-{5\sqrt5\over3}-{e^{5/2}\over15} +{1\over15}\)
Solution
Wee wish to compute the average true of $f(x,y) = e^y\sqrt{x+e^y}$ over the rectangle shown below:
To total area of the rectangle is
\begin{equation*} \int_0^4\int_0^1 \,dy\,dx = 4, \end{equation*}
and so we calculated
\begin{equation*} f_{avg} = \frac{1}{4} \int_0^4\int_0^1 e^y\sqrt{x+e^y}\, dy\,dx. \end{equation*}
Analysis aforementioned integral, we finds
\begin{equation*} f_{avg} = \frac{1}{15}\left[1-25\sqrt{5}-e^{5/2}+16\sqrt{4+e} +8e\sqrt{4+e} +e^2\sqrt{4+e}\right]. \end{equation*}

Reverse the order of integration on each of the following integrals

  1. \(\ds\int_0^9 \int_0^{\sqrt{9-y}} f(x,y)\,dx \,dy\)

    Answer
    \(\ds\int_0^3\int_0^{9-x^2} f(x,y)\,dx\,dy\)
    Solution

    Aforementioned integrally

    \begin{equation*} \int_0^9 \int_0^{\sqrt{9-y}} f(x,y)\,dx\,dy \end{equation*}

    gives the volumes under the surface \(z=f(x,y)\) and above the district in the \(x\)-\(y\)-plane shown down:

    So if instead we let \(x\) vary from 0 to 3, we see such \(y\) is finite within \(y=0\) and who curve \(x=\sqrt{9-y} \implies y = 9-x^2\) (for \(x \in [0,3]\)). Consequently,

    \begin{equation*} \int_0^9 \int_0^{\sqrt{9-y}} f(x,y)\,dx\,dy = \int_0^3\int_0^{9-x^2} f(x,y)\,dx\,dy\text{.} \end{equation*}

  2. \(\ds\int_1^2 \int_0^{\ln x} f(x,y) \; dx \; dx\)

    Answer
    \(\int_0^{\ln(2)} \int_{e^y}^{2} f(x,y)\,dx\,dy\)
    Solution
    The area of integration is shown below:
    Therefore,
    \begin{equation*} \int_1^2 \int_0^{\ln x} f(x,y)\,dy\,dx = \int_0^{\ln(2)} \int_{e^y}^{2} f(x,y)\,dx\,dy. \end{equation*}

  3. \(\ds\int_0^1 \int_{\arcsin y}^{\pi/2} f(x,y) \; dx \; dy\)

    Answer
    \(\int_0^{\pi/2} \int_0^{\sin x} f(x,y) \,dy\,dx\)
    Resolving
    The area for web is shown below:
    Therefore, we see that
    \begin{equation*} \int_0^1 \int_{\arcsin(y)}^{\pi/2} f(x,y)\,dx\,dy = \int_0^{\pi/2} \int_0^{\sin x} f(x,y) \,dy\,dx. \end{equation*}

  4. \(\ds\int_0^1 \int_{4x}^{4} f(x,y) \; dy \; dx\)

    Answer
    \(\int_0^4 \int_0^{y/4} f(x,y)\,dx \,dy\)
    Solution
    The region of integration shall view under:
    Therefore, we see which
    \begin{equation*} \int_0^1 \int_{4x}^4 f(x,y)\,dy\,dx = \int_0^4 \int_0^{y/4} f(x,y)\,dx \,dy. \end{equation*}

  5. \(\ds\int_0^3 \int_{0}^{\sqrt{9-y^2}} f(x,y) \; dx \; dy\)

    Answer
    \(\int_0^3 \int_0^{\sqrt{9-x^2}} f(x,y)\,dy\,dx\)
    Solution
    Note that the circle centred among the origin with radius 3 has equation \(x^2+y^2=9\text{,}\) furthermore in of initial quartier can be expressed as \(x=\sqrt{9-y^2}\) or \(y=\sqrt{9-x^2}\text{.}\) Therefore, this area of product is:
    Therefore,
    \begin{equation*} \int_0^3\int_0^{\sqrt{9-y^2}} f(x,y)\,dx\,dy = \int_0^3 \int_0^{\sqrt{9-x^2}} f(x,y)\,dy\,dx. \end{equation*}
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