Section 13.3 Double Integration with Polar Coordinates
We have used iterated integrals to evaluate double integrals, which give the signed volume under a surface, z=f(x,y),z=f(x,y), over a region RR of the xyxy-plane. The integrand is simply f(x,y),f(x,y), and the bounds of the integrals are determined by the region R.R. Some regions RR are easy to describe using rectangular coordinates — that is, with equations of the form y=f(x),y=f(x), x=a,x=a, etc. However, some regions are easier to handle if we represent their boundaries with polar equations of the form r=f(θ),r=f(θ), θ=α,θ=α, etc.Key Idea 13.3.3. Evaluating Double Integrals with Polar Coordinates.
Let z=f(x,y) be a continuous function defined over a closed, bounded region R in the xy-plane, where R is bounded by the polar equations α≤θ≤β and g1(θ)≤r≤g2(θ). Then
Example 13.3.4. Evaluating a double integral with polar coordinates.
Find the signed volume under the plane z=4−x−2y over the disk bounded by the circle with equation x2+y2=1.
The bounds of the integral are determined solely by the region \(R\) over which we are integrating. In this case, it is a disk with boundary \(x^2+y^2=1\text{.}\) We need to find polar bounds for this region. It may help to review Section 9.4; bounds for this disk are \(0\leq r\leq 1\) and \(0\leq \theta\leq 2\pi\text{.}\)
We replace \(f(x,y)\) with \(f(r\cos(\theta) ,r\sin(\theta) )\text{.}\) That means we make the following substitutions:
Finally, we replace \(dA\) in the double integral with \(r\, dr\, d\theta\text{.}\) This gives the final iterated integral, which we evaluate:
The surface and region \(R\) are shown in Figure 13.3.5.
Example 13.3.7. Evaluating a double integral with polar coordinates.
Find the volume under the paraboloid z=4−(x−2)2−y2 over the region bounded by the circles (x−1)2+y2=1 and (x−2)2+y2=4.
At first glance, this seems like a very hard volume to compute as the region \(R\) (shown in Figure 13.3.8.(a)) has a hole in it, cutting out a strange portion of the surface, as shown in Figure 13.3.8.(b). However, by describing \(R\) in terms of polar equations, the volume is not very difficult to compute.
It is straightforward to show that the circle \((x-1)^2+y^2=1\) has polar equation \(r=2\cos(\theta)\text{,}\) and that the circle \((x-2)^2+y^2=4\) has polar equation \(r=4\cos(\theta)\text{.}\) Each of these circles is traced out on the interval \(0\leq\theta\leq\pi\text{.}\) The bounds on \(r\) are \(2\cos(\theta) \leq r\leq 4\cos(\theta)\text{.}\)
Replacing \(x\) with \(r\cos(\theta)\) in the integrand, along with replacing \(y\) with \(r\sin(\theta)\text{,}\) prepares us to evaluate the double integral \(\iint_Rf(x,y)\, dA\text{:}\)
To integrate \(\cos^4(\theta)\text{,}\) rewrite it as \(\cos^2(\theta) \cos^2(\theta)\) and employ the power-reducing formula twice:
Picking up from where we left off above, we have
While this example was not trivial, the double integral would have been much harder to evaluate had we used rectangular coordinates.
Example 13.3.9. Evaluating a double integral with polar coordinates.
Find the volume under the surface f(x,y)=1x2+y2+1 over the sector of the circle with radius a centered at the origin in the first quadrant, as shown in Figure 13.3.10.
The region \(R\) we are integrating over is a circle with radius \(a\text{,}\) restricted to the first quadrant. Thus, in polar, the bounds on \(R\) are \(0\leq r\leq a\text{,}\) \(0\leq\theta\leq\pi/2\text{.}\) The integrand is rewritten in polar as
We find the volume as follows:
Figure 13.3.10 shows that \(f\) shrinks to near 0 very quickly. Regardless, as \(a\) grows, so does the volume, without bound.
Example 13.3.11. Finding the volume of a sphere.
Find the volume of a sphere with radius a.
The sphere of radius \(a\text{,}\) centered at the origin, has equation \(x^2+y^2+z^2=a^2\text{;}\) solving for \(z\text{,}\) we have \(z=\sqrt{a^2-x^2-y^2}\text{.}\) This gives the upper half of a sphere. We wish to find the volume under this top half, then double it to find the total volume.
The region we need to integrate over is the disk of radius \(a\text{,}\) centered at the origin. Polar bounds for this equation are \(0\leq r\leq a\text{,}\) \(0\leq\theta\leq2\pi\text{.}\)
All together, the volume of a sphere with radius \(a\) is:
We can evaluate this inner integral with substitution. With \(u=a^2-r^2\text{,}\) \(du = -2r\, dr\text{.}\) The new bounds of integration are \(u(0) = a^2\) to \(u(a)=0\text{.}\) Thus we have:
\begin{align*} \amp = \int_0^{2\pi}\int_{a^2}^0\big(-u^{1/2}\big)\, du\, d\theta\\ \amp = \int_0^{2\pi}\left.\left(-\frac23u^{3/2}\right)\right|_{a^2}^0 d\theta\\ \amp = \int_0^{2\pi}\left(\frac23a^3\right)\, d\theta\\ \amp = \left.\left(\frac23a^3\theta\right)\right|_0^{2\pi}\\ \amp = \frac43\pi a^3\text{.} \end{align*}Generally, the formula for the volume of a sphere with radius \(r\) is given as \(4/3\pi r^3\text{;}\) we have justified this formula with our calculation.
Example 13.3.13. Finding the volume of a solid.
A sculptor wants to make a solid bronze cast of the solid shown in Figure 13.3.14, where the base of the solid has boundary, in polar coordinates, r=cos(3θ), and the top is defined by the plane z=1−x+0.1y. Find the volume of the solid.
From the outset, we should recognize that knowing how to set up this problem is probably more important than knowing how to compute the integrals. The iterated integral to come is not “hard” to evaluate, though it is long, requiring lots of algebra. Once the proper iterated integral is determined, one can use readily available technology to help compute the final answer.
The region \(R\) that we are integrating over is bound by \(0\leq r\leq \cos(3\theta)\text{,}\) for \(0\leq \theta\leq\pi\) (note that this rose curve is traced out on the interval \([0,\pi]\text{,}\) not \([0,2\pi]\)). This gives us our bounds of integration. The integrand is \(z=1-x+0.1y\text{;}\) converting to polar, we have that the volume \(V\) is:
Distributing the \(r\text{,}\) the inner integral is easy to evaluate, leading to
This integral takes time to compute by hand; it is rather long and cumbersome. The powers of cosine need to be reduced, and products like \(\cos(3\theta)\cos(\theta)\) need to be turned to sums using the Product To Sum formulas in the back cover of this text.
We rewrite \(\frac12\cos^2(3\theta)\) as \(\frac14(1+\cos(6\theta))\text{.}\) We can also rewrite \(\frac13\cos^3(3\theta)\cos(\theta)\) as:
This last expression still needs simplification, but eventually all terms can be reduced to the form \(a\cos(m\theta)\) or \(a\sin(m\theta)\) for various values of \(a\) and \(m\text{.}\)
We forgo the algebra and recommend the reader employ technology, such as WolframAlpha\textregistered, to compute the numeric answer. Such technology gives:
Since the units were not specified, we leave the result as almost \(0.8\) cubic units (meters, feet, etc.) Should the artist want to scale the piece uniformly, so that each rose petal had a length other than 1, she should keep in mind that scaling by a factor of \(k\) scales the volume by a factor of \(k^3\text{.}\)
Exercises Exercises
Terms and Concepts
In the following exercises, a function f(x,y) is given and a region R of the xy-plane is described. Set up and evaluate ∬Rf(x,y)dA using polar coordinates.
3.
Use polar coordinates to evaluate ∬Rf(x,y)dA, where f(x,y)=3x−y+4 and R is the region enclosed by the circle x2+y2=1.
4.
f(x,y)=4x+4y; R is the region enclosed by the circle x2+y2=4.
5.
Use polar coordinates to evaluate ∬Rf(x,y)dA, where f(x,y)=8−y and R is the region enclosed by the circles with polar equations r=cos(θ) and r=3cos(θ).
6.
f(x,y)=4; R is the region enclosed by the petal of the rose curve r=sin(2θ) in the first quadrant.
7.
f(x,y)=ln(x2+y2); R is the annulus enclosed by the circles x2+y2=1 and x2+y2=4.
8.
Use polar coordinates to evaluate ∬Rf(x,y)dA, where f(x,y)=1−x2−y2 and R is the region enclosed by the circle x2+y2=1.
9.
f(x,y)=x2−y2; R is the region enclosed by the circle x2+y2=36 in the first and fourth quadrants.
10.
f(x,y)=(x−y)/(x+y); R is the region enclosed by the lines y=x, y=0 and the circle x2+y2=1 in the first quadrant.
In the following exercises, an iterated integral in rectangular coordinates is given. Rewrite the integral using polar coordinates and evaluate the new double integral.
11.
∫50∫√25−x2−√25−x2√x2+y2dydx
12.
Evalueate the integral ∫4−4∫0−√16−y2(2y−x)dxdy by first rewriting it using polar coordinates.
13.
∫20∫√8−y2y(x+y)dxdy
14.
∫−1−2∫√4−x20(x+5)dydx+∫1−1∫√4−x2√1−x2(x+5)dydx+∫21∫√4−x20(x+5)dydx
Hint: draw the region of each integral carefully and see how they all connect.
In the following exercises, special double integrals are presented that are especially well suited for evaluation in polar coordinates.
15.
Consider ∬Re−(x2+y2)dA.
Why is this integral difficult to evaluate in rectangular coordinates, regardless of the region R?
Let R be the region bounded by the circle of radius a centered at the origin. Evaluate the double integral using polar coordinates.
Take the limit of your answer from (b), as a→∞. What does this imply about the volume under the surface of e−(x2+y2) over the entire xy-plane?
16.
The surface of a right circular cone with height h and base radius a can be described by the equation f(x,y)=h−h√x2a2+y2a2, where the tip of the cone lies at (0,0,h) and the circular base lies in the xy-plane, centered at the origin.
Confirm that the volume of a right circular cone with height h and base radius a is V=13πa2h by evaluating ∬Rf(x,y)dA in polar coordinates.