(Problem 1a) Compute where is the region between the circles and .
In this section we compute double integrals using polar coordinates.
Integrating in polar coordinates requires knowledge of the area of a polar rectangle. A polar rectangle is a region in the -plane has the form
Its area is computed as the difference of two sectors: \begin{align*} S_1&: 0 \leq r\leq r_1, \; \theta _1 \leq \theta \leq \theta _2 \quad \text{and} \\ S_2&: 0 \leq r\leq r_2, \; \theta _1 \leq \theta \leq \theta _2 \end{align*}
and is given by \begin{align*} A &= \frac 12 (\theta _2 - \theta _1) r_2^2 - \frac 12 (\theta _2 -\theta _1) r_1^2\\ &= \frac 12 (\theta _2 - \theta _1) (r_2^2 - r_1^2)\\ &= \frac 12 (\theta _2 - \theta _1) (r_2 - r_1)(r_2 + r_1)\\ &= \frac 12(r_2 + r_1) \Delta \theta \Delta r \\ &= r^* \Delta r \Delta \theta \end{align*}
where See the figure below.
Suppose that the region is a polar rectangle. When converting a double integral over into polar coordinates, the formula for the area of a polar rectangle suggests that
Example 1 Compute where is the quarter circle in the first quadrant bounded by
and
Writing this integral as iterated integrals will leave us with a difficult anti-differentiation
problem regardless of the order chosen. However, the integral is fairly easy to
compute using polar coordinates. The region can be seen as a polar rectangle
with Moreover, the function transforms into a simpler function in polar
coordinates. With the function becomes since . Using polar coordinates, we have \begin{align*} \iint _R \sqrt{x^2 + y^2} \, dA & = \int _0^{\pi /2} \int _0^2 r \cdot r \, dr \, d\theta \\ & = \int _0^{\pi /2} \frac{r^3}{3} \bigg |_{0}^2 \, d\theta \\ & = \int _0^{\pi /2} \frac 83\, d\theta \\ &= \frac{4\pi }{3} \end{align*}
Here is a video solution of problems 1a and 1b:
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Example 2 Compute where is polar rectangle and .
Since and , we have Since, , In polar coordinates, the integral becomes \begin{align*} \iint _R \tan ^{-1}\left (\frac{y}{x}\right ) \, dA & = \int _{-\pi /2}^{\pi /2} \int _2^3 \theta \cdot r \, dr \, d\theta \\ & = \int _{-\pi /2}^{\pi /2} \theta \cdot \frac{r^2}{2}\bigg |_2^3 \, d\theta \\ & = \frac 52 \int _{-\pi /2}^{\pi /2} \theta \, d\theta \\ &= \frac 52 \frac{\theta ^2}{2} \bigg |_{-\pi /2}^{\pi /2}\\ &= \frac 52 \left (\frac{\pi ^2}{4} - \frac{\pi ^2}{4}\right )\\ &= 0 \end{align*}
Here is a video solution of problems 2a and 2b:
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Example 3 Find the volume of the region below the paraboloid and above the
-plane.
\begin{align*} \iint _R (1-x^2 - y^2) \, dA &= \int _0^{2\pi } \int _0^1 (1-r^2) \cdot r \, dr \, d\theta \\ &= 2\pi \left (\frac{r^2}{2} - \frac{r^4}{4}\right ) \bigg |_0^1\\ &= 2\pi \cdot \frac 14 = \frac{\pi }{2} \end{align*}
Here is a video solution of problem 3:
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Example 4 Compute the doubly improper integral There is no known antiderivative
for the function (without resorting to power series), but there is a way to compute
this integral using polar coordinates. Recall that if a function is separable,
then we can write a double integral over a rectangle as a product of definite
integrals: Our tactic is to compute the square of the original integral, written
in the form Recombining these integrals into iterated integrals (using the
fact that ) we have Applying the polar coordinate transformation yields \begin{align*} \left (\int _{-\infty }^{\infty } e^{-x^2} \, dx \right )^2 &= \int _{-\infty }^{\infty }\int _{-\infty }^{\infty } e^{-x^2 - y^2} \, dy\, dx \\ &= \int _0^{2\pi } \int _0^\infty e^{-r^2} \cdot r \, dr \, d\theta \\ &= 2\pi \int _0^\infty r e^{-r^2} \, dr\\ &= -\pi \int _0^{-\infty } e^u \, du\\ &= \pi \int _{-\infty }^0 e^u \, du\\ &= \pi \lim _{t \to -\infty } \int _t^0 e^u \, du\\ &= \pi \lim _{t \to -\infty } \left (1 - e^t\right )\\ &= \pi (1-0) = \pi \end{align*}
Thus, for the original integral, we have
Here is a video solution of problem 4:
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1 The Jacobian of a Transformation
In the double integral the area element is equal to when the double integral is written as iterated integrals. We noted that if we apply a polar coordinate transformation to the variables and to obtain the variables and , then the area element in the double integral is In general, when and are transformed into two other variables, and , the area element in the double integral becomes where is the Jacobian of the transformation, defined by the following matrix determinant:
The Jacobian can also be extended into higher dimensions, as we shall see when we
study triple integrals in spherical coordinates.
Example 5 Let and , i.e., the polar coordinate transformation. Compute the
Jacobian, and the corresponding area element, .
The Jacobian is given by the determinant of the matrix: \begin{align*} \begin{vmatrix} \pp [x]{r} & \pp [x]{\theta }\\[10pt] \pp [y]{r} & \pp [y]{\theta }\\ \end{vmatrix} &= \begin{vmatrix} \cos \theta & -r\sin \theta \\[10pt] \sin \theta & r\cos \theta \\ \end{vmatrix}\\ &= r \cos ^2 \theta + r \sin ^2 \theta \\ &= r \end{align*}
The Jacobian is given by the determinant of the matrix: \begin{align*} \begin{vmatrix} \pp [x]{r} & \pp [x]{\theta }\\[10pt] \pp [y]{r} & \pp [y]{\theta }\\ \end{vmatrix} &= \begin{vmatrix} \cos \theta & -r\sin \theta \\[10pt] \sin \theta & r\cos \theta \\ \end{vmatrix}\\ &= r \cos ^2 \theta + r \sin ^2 \theta \\ &= r \end{align*}
Hence, the area element is The last equality follows from the fact that .
Here is a video solution of problem 5c:
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