Recognize the format of a twin integral over a polar rectangle-shaped region. Advice a twin integral in polar collaborates by making use of an iterated integral. Identify the layout of a twin integral over a general polar region. Use double integrals in polar collaborates to calculate areas and volumes.

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Double integrals are occasionally much easier to advice if we adjust rectangular coordinates to polar coordinates. However, prior to we describe how to do this change, we require to establish the principle of a double integral in a polar rectangle-shaped region.

## Polar rectangular Regions the Integration

When we defined the double integral because that a continuous function in rectangular coordinates—say, $$g$$ over a region $$R$$ in the $$xy$$-plane—we split $$R$$ into subrectangles through sides parallel to the name: coordinates axes. This sides have either constant $$x$$-values and/or constant $$y$$-values. In polar coordinates, the shape we work with is a polar rectangle, whose political parties have constant $$r$$-values and/or consistent $$\theta$$-values. This method we can explain a polar rectangle as in figure $$\PageIndex1a$$, with $$R = \(r,\theta)\,$$.

Figure $$\PageIndex1$$: (a) A polar rectangle $$R$$ (b) divided into subrectangles $$R_ij$$ (c) Close-up the a subrectangle.

In this section, we are looking to incorporate over polar rectangles. Think about a role $$f(r,\theta)$$ over a polar rectangle $$R$$. We division the expression  right into $$m$$ subintervals  of size $$\Delta r = (b - a)/m$$ and also divide the expression $$<\alpha, \beta>$$ into $$n$$ subintervals $$<\theta_i-1, \theta_i>$$ of broad $$\Delta \theta = (\beta - \alpha)/n$$. This means that the one $$r = r_i$$ and rays $$\theta = \theta_i$$ for $$1 \leq ns \leq m$$ and also $$1 \leq j \leq n$$ divide the polar rectangle $$R$$ into smaller polar subrectangles $$R_ij$$ (Figure $$\PageIndex1b$$).

As before, we require to find the area $$\Delta A$$ the the polar subrectangle $$R_ij$$ and also the “polar” volume the the thin box over $$R_ij$$. Recall that, in a one of radius $$r$$ the size $$s$$ of an arc subtended by a central angle of $$\theta$$ radians is $$s = r\theta$$. Notification that the polar rectangle $$R_ij$$ looks a lot favor a trapezoid through parallel political parties $$r_i-1\Delta \theta$$ and $$r_i\Delta \theta$$ and also with a width $$\Delta r$$. Thus the area of the polar subrectangle $$R_ij$$ is

\<\Delta A = \frac12 \Delta r (r_i-1 \Delta \theta + r_i \Delta \theta ). \nonumber\>

Simplifying and letting

\

we have actually $$\Delta A = r_ij^* \Delta r \Delta \theta$$.

Therefore, the polar volume that the thin box above $$R_ij$$ (Figure $$\PageIndex2$$) is

\

Figure $$\PageIndex2$$: detect the volume the the slim box above polar rectangle $$R_ij$$.

Using the exact same idea for all the subrectangles and also summing the quantities of the rectangular boxes, we achieve a double Riemann amount as

\<\sum_i=1^m \sum_j=1^n f(r_ij^*, \theta_ij^*) r_ij^* \Delta r \Delta \theta. \nonumber\>

As we have actually seen before, we obtain a better approximation to the polar volume the the solid over the region $$R$$ as soon as we allow $$m$$ and also $$n$$ come to be larger. Hence, we define the polar volume together the limit of the double Riemann sum,

\

This becomes the expression for the twin integral.

Again, simply as in section on double Integrals over rectangular Regions, the dual integral over a polar rectangular region can it is in expressed as an iterated integral in polar coordinates. Hence,

\<\iint_R f(r, \theta)\,dA = \iint_R f(r, \theta) \,r \, dr \, d\theta = \int_\theta=\alpha^\theta=\beta \int_r=a^r=b f(r,\theta) \,r \, dr \, d\theta.\>

Notice the the expression because that $$dA$$ is replaced by $$r \, dr \, d\theta$$ when working in polar coordinates. Another way to look in ~ the polar twin integral is to change the double integral in rectangular collaborates by substitution. As soon as the duty $$f$$ is provided in regards to $$x$$ and $$y$$ utilizing $$x = r \, \cos \, \theta, \, y = r \, \sin \, \theta$$, and $$dA = r \, dr \, d\theta$$ alters it to

\<\iint_R f(x,y) \,dA = \iint_R f(r \, \cos \, \theta, \, r \, \sin \, \theta ) \,r \, dr \, d\theta.\>

Note the all the properties noted in ar on dual Integrals over rectangle-shaped Regions for the twin integral in rectangular collaborates hold true because that the dual integral in polar works with as well, so we have the right to use them without hesitation.

Example $$\PageIndex1A$$: Sketching a Polar rectangular Region

Sketch the polar rectangle-shaped region

\

Solution

As we can see from number $$\PageIndex3$$, $$r = 1$$ and also $$r = 3$$ are circles that radius 1 and 3 and also $$0 \leq \theta \leq \pi$$ covers the whole top fifty percent of the plane. For this reason the an ar $$R$$ looks prefer a semicircular band.

Figure $$\PageIndex3$$: The polar region $$R$$ lies in between two semicircles.

Now that we have actually sketched a polar rectangular region, permit us demonstrate how to evaluate a double integral end this region by utilizing polar coordinates.

Example $$\PageIndex1B$$: assessing a twin Integral end a Polar rectangle-shaped Region

Evaluate the integral $$\displaystyle \iint_R 3x \, dA$$ over the region $$R = \\,1 \leq r \leq 2, \, 0 \leq \theta \leq \pi \.$$

Solution

First we sketch a figure comparable to figure $$\PageIndex3$$, yet with outer radius $$r=2$$. Indigenous the figure we have the right to see that we have

\<\beginalign* \iint_R 3x \, dA &= \int_\theta=0^\theta=\pi \int_r=1^r=2 3r \, \cos \, \theta \,r \, dr \, d\theta \quad\textUse one integral with correct boundaries of integration. \\ &= \int_\theta=0^\theta=\pi \cos \, \theta \left<\left. R^3\right|_r=1^r=2\right> d\theta \quad\textIntegrate first with respect come $r$. \\ &=\int_\theta=0^\theta=\pi 7 \, \cos \, \theta \, d\theta \\ &= 7 \, \sin \, \theta \bigg|_\theta=0^\theta=\pi = 0. \endalign*\>

Exercise $$\PageIndex1$$

Sketch the region $$D = \ (r,\theta) \vert 1\leq r \leq 2, \, -\frac\pi2 \leq \theta \leq \frac\pi2 \$$, and also evaluate $$\displaystyle \iint_R x \, dA$$.

Hint

Follow the steps in instance $$\PageIndex1A$$.

$$\frac143$$

Example $$\PageIndex2A$$: analyzing a twin Integral by convert from rectangular Coordinates

Evaluate the integral

\<\iint_R (1 - x^2 - y^2) \,dA \nonumber\>

where $$R$$ is the unit circle on the $$xy$$-plane.

Solution

The region $$R$$ is a unit circle, so us can describe it together $$R = \\,0 \leq r \leq 1, \, 0 \leq \theta \leq 2\pi \$$.

Using the conversion $$x = r \, \cos \, \theta, \, y = r \, \sin \, \theta$$, and also $$dA = r \, dr \, d\theta$$, us have

\<\beginalign* \iint_R (1 - x^2 - y^2) \,dA &= \int_0^2\pi \int_0^1 (1 - r^2) \,r \, dr \, d\theta \\<4pt> &= \int_0^2\pi \int_0^1 (r - r^3) \,dr \, d\theta \\ &= \int_0^2\pi \left<\fracr^22 - \fracr^44\right>_0^1 \,d\theta \\&= \int_0^2\pi \frac14\,d\theta = \frac\pi2. \endalign*\>

Example $$\PageIndex2B$$: assessing a twin Integral by converting from rectangle-shaped Coordinates

Evaluate the integral \<\displaystyle \iint_R (x + y) \,dA \nonumber\> where $$R = \big\\,1 \leq x^2 + y^2 \leq 4, \, x \leq 0 \big\.$$

Solution

We have the right to see the $$R$$ is one annular an ar that can be convert to polar coordinates and also described together $$R = \left\(r, \theta)\,$$ (see the adhering to graph).

Figure $$\PageIndex4$$: The annular an ar of integration $$R$$.

Hence, using the conversion $$x = r \, \cos \, \theta, \, y = r \, \sin \, \theta$$, and $$dA = r \, dr \, d\theta$$, we have

\<\beginalign* \iint_R (x + y)\,dA &= \int_\theta=\pi/2^\theta=3\pi/2 \int_r=1^r=2 (r \, \cos \, \theta + r \, \sin \, \theta) r \, dr \, d\theta \\ &= \left(\int_r=1^r=2 r^2 \, dr\right)\left(\int_\pi/2^3\pi/2 (\cos \, \theta + \sin \, \theta)\,d\theta\right) \\ &= \left. \left<\fracr^33\right>_1^2 <\sin \, \theta - \cos \, \theta> \right|_\pi/2^3\pi/2 \\ &= - \frac143. \endalign*\>

## General Polar regions of Integration

To evaluate the twin integral the a consistent function by iterated integrals over basic polar regions, we consider two types of regions, analogous to kind I and form II as debated for rectangular collaborates in ar on dual Integrals over basic Regions. The is an ext common to create polar equations as $$r = f(\theta)$$ than $$\theta = f(r)$$, therefore we describe a general polar region as $$R = \(r, \theta)\,$$ (Figure $$\PageIndex5$$).

api/deki/files/10968/imageedit_7_8938721574.png?revision=1" />Figure $$\PageIndex6$$: The region $$D$$ is the top fifty percent of a cardioid.

Hence, us have

\<\beginalign* \iint_D r^2 \sin \, \theta \, r \, dr \, d\theta &= \int_\theta=0^\theta=\pi \int_r=0^r=1+\cos \theta (r^2 \sin \, \theta) \,r \, dr \, d\theta \\ &= \frac14\left.\int_\theta=0^\theta=\pi \right|_r=0^r=1+\cos \, \theta \sin \, \theta \, d\theta \\ &= \frac14 \int_\theta=0^\theta=\pi (1 + \cos \, \theta )^4 \sin \, \theta \, d\theta \\ &= - \frac14 \left< \frac(1 + \cos \, \theta)^55\right>_0^\pi = \frac85.\endalign*\>

Exercise $$\PageIndex3$$

Evaluate the integral

\<\iint_D r^2 \sin^2 2\theta \,r \, dr \, d\theta \nonumber\>

where $$D = \left\\,0 \leq \theta \leq \pi, \, 0 \leq r \leq 2 \sqrt\cos \, 2\theta \right\$$.

Hint

Graph the an ar and monitor the actions in the previous example.

$$\frac\pi8$$

## Polar Areas and Volumes

As in rectangular coordinates, if a heavy $$S$$ is bounded through the surface $$z = f(r, \theta)$$, and by the surface $$r = a, \, r = b, \, \theta = \alpha$$, and also $$\theta = \beta$$, we can uncover the volume $$V$$ of $$S$$ by twin integration, as

\

If the basic of the solid have the right to be explained as $$D = \(r, \theta)$$, climate the double integral for the volume becomes

\

We highlight this idea v some examples.

Example $$\PageIndex4A$$: finding a Volume making use of a double Integral

Find the volume the the solid the lies under the paraboloid $$z = 1 - x^2 - y^2$$ and above the unit one on the $$xy$$-plane (Figure $$\PageIndex7$$).

Figure $$\PageIndex7$$: finding the volume of a solid under a paraboloid and over the unit circle.

Solution

By the method of twin integration, we can see that the volume is the iterated integral that the form

\<\displaystyle \iint_R (1 - x^2 - y^2)\,dA \nonumber\>

where $$R = \big\\,0 \leq r \leq 1, \, 0 \leq \theta \leq 2\pi\big\$$.

This integration to be shown before in example $$\PageIndex2A$$, for this reason the volume is $$\frac\pi2$$ cubic units.

Example $$\PageIndex4B$$: detect a Volume Using dual Integration

Find the volume the the solid that lies under the paraboloid $$z = 4 - x^2 - y^2$$ and above the disc $$(x - 1)^2 + y^2 = 1$$ on the $$xy$$-plane. See the paraboloid in number $$\PageIndex8$$ intersecting the cylinder $$(x - 1)^2 + y^2 = 1$$ above the $$xy$$-plane.

Figure $$\PageIndex8$$: recognize the volume that a solid v a paraboloid cap and also a one base.

Solution

First change the decaying $$(x - 1)^2 + y^2 = 1$$ come polar coordinates. Broadening the square term, we have actually $$x^2 - 2x + 1 + y^2 = 1$$. Then simplify to gain $$x^2 + y^2 = 2x$$, which in polar coordinates becomes $$r^2 = 2r \, \cos \, \theta$$ and then either $$r = 0$$ or $$r = 2 \, \cos \, \theta$$. Similarly, the equation of the paraboloid alters to $$z = 4 - r^2$$. As such we can define the disc $$(x - 1)^2 + y^2 = 1$$ top top the $$xy$$ -plane together the region

\

Hence the volume of the heavy bounded above by the paraboloid $$z = 4 - x^2 - y^2$$ and also below through $$r = 2 \, \cos \theta$$ is

\<\beginalign* V &= \iint_D f(r, \theta) \,r \, dr \, d\theta \\&= \int_\theta=0^\theta=\pi \int_r=0^r=2 \, \cos \, \theta (4 - r^2) \,r \, dr \, d\theta\\ &= \int_\theta=0^\theta=\pi\left.\left<4\fracr^22 - \fracr^44\right|_0^2 \, \cos \, \theta\right>d\theta \\ &= \int_0^\pi <8 \, \cos^2\theta - 4 \, \cos^4\theta>\,d\theta \\&= \left<\frac52\theta + \frac52 \sin \, \theta \, \cos \, \theta - \sin \, \theta \cos^3\theta \right>_0^\pi = \frac52\pi\; \textunits^3. \endalign*\>

Example $$\PageIndex5A$$: finding a Volume making use of a double Integral

Find the volume the the an ar that lies under the paraboloid $$z = x^2 + y^2$$ and over the triangle fastened by the currently $$y = x, \, x = 0$$, and $$x + y = 2$$ in the $$xy$$-plane.

Solution

First examine the region over i beg your pardon we need to collection up the double integral and also the accompanying paraboloid.

Figure $$\PageIndex9$$: recognize the volume that a heavy under a paraboloid and above a provided triangle.

The region $$D$$ is $$\(x,y)\,$$. Convert the currently $$y = x, \, x = 0$$, and also $$x + y = 2$$ in the $$xy$$-plane to functions of $$r$$ and also $$\theta$$ we have $$\theta = \pi/4, \, \theta = \pi/2$$, and $$r = 2 / (\cos \, \theta + \sin \, \theta)$$, respectively. Graphing the an ar on the $$xy$$- plane, we watch that that looks choose $$D = \\,\pi/4 \leq \theta \leq \pi/2, \, 0 \leq r \leq 2/(\cos \, \theta + \sin \, \theta)\$$.

Now converting the equation that the surface provides $$z = x^2 + y^2 = r^2$$. Therefore, the volume that the heavy is provided by the twin integral

\<\beginalign* V &= \iint_D f(r, \theta)\,r \, dr \, d\theta \\&= \int_\theta=\pi/4^\theta=\pi/2 \int_r=0^r=2/ (\cos \, \theta + \sin \, \theta) r^2 r \, dr d\theta \\ &= \int_\pi/4^\pi/2\left<\fracr^44\right>_0^2/(\cos \, \theta + \sin \, \theta) d\theta \\ &=\frac14\int_\pi/4^\pi/2 \left(\frac2\cos \, \theta + \sin \, \theta\right)^4 d\theta \\ &= \frac164 \int_\pi/4^\pi/2 \left(\frac1\cos \, \theta + \sin \, \theta \right)^4 d\theta \\&= 4\int_\pi/4^\pi/2 \left(\frac1\cos \, \theta + \sin \, \theta\right)^4 d\theta. \endalign*\>

As you deserve to see, this integral is very complicated. So, we deserve to instead advice this double integral in rectangular works with as

\

Evaluating gives

\<\beginalign* V &= \int_0^1 \int_x^2-x (x^2 + y^2) \,dy \, dx \\&= \int_0^1 \left.\left\right|_x^2-x dx\\ &= \int_0^1 \frac83 - 4x + 4x^2 - \frac8x^33 \,dx \\ &= \left.\left<\frac8x3 - 2x^2 + \frac4x^33 - \frac2x^43\right>\right|_0^1 \\&= \frac43 \; \textunits^3. \endalign*\>

Example $$\PageIndex5B$$: recognize a Volume making use of a double Integral

Use polar coordinates to discover the volume inside the cone $$z = 2 - \sqrtx^2 + y^2$$ and above the $$xy$$-plane.

Solution

The an ar $$D$$ for the integration is the base of the cone, which shows up to it is in a circle on the $$xy$$-plane (Figure $$\PageIndex10$$).

api/deki/files/10974/imageedit_22_9774650897.png?revision=1" />Figure $$\PageIndex11$$: detect the area the a polar climbed with eight petals.

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Using symmetry, we deserve to see the we need to uncover the area that one petal and then main point it by 8. An alert that the worths of $$\theta$$ for which the graph passes with the beginning are the zeros that the role $$\cos \, 4\theta$$, and these space odd multiples of $$\pi/8$$. Thus, one of the petals synchronizes to the worths of $$\theta$$ in the expression $$<-\pi/8, \pi/8>$$. Therefore, the area bounded through the curve $$r = \cos \, 4\theta$$ is

\<\beginalign* A &= 8 \int_\theta=-\pi/8^\theta=\pi/8 \int_r=0^r=\cos \, 4\theta 1\,r \, dr \, d\theta \\ &= 8 \int_\theta=-\pi/8^\theta=\pi/8\left.\left<\frac12r^2\right|_0^\cos \, 4\theta\right> d\theta \\ &= 8 \int_-\pi/8^\pi/8 \frac12 \cos^24\theta \, d\theta \\&= 8\left. \left<\frac14 \theta + \frac116 \sin \, 4\theta \, \cos \, 4\theta \right|_-\pi/8^\pi/8\right> \\&= 8 \left<\frac\pi16\right> = \frac\pi2\; \textunits^2. \endalign*\>

api/deki/files/10973/imageedit_25_4423207893.png?revision=1" />Figure $$\PageIndex12$$: detect the area enclosed by both a circle and a cardioid.

We have the right to from check out the the opposite of the graph that we require to uncover the points of intersection. Setting the two equations same to each various other gives

\<3 \, \cos \, \theta = 1 + \cos \, \theta. \nonumber\>

One of the points of intersection is $$\theta = \pi/3$$. The area above the polar axis consists of two parts, v one part defined through the cardioid indigenous $$\theta = 0$$ come $$\theta = \pi/3$$ and the other part defined through the circle from $$\theta = \pi/3$$ come $$\theta = \pi/2$$. Through symmetry, the full area is twice the area over the polar axis. Thus, us have