5.3 Double integrals in polar coordinates (Page 3/7)

Calculus volume 3 Page 3 / 7

Evaluating a double integral over a general polar region

Evaluate the integral $\iint_{D} r^{2} sin θ r d r d θ$ where $D$ is the region bounded by the polar axis and the upper half of the cardioid $r = 1 + cos θ .$

We can describe the region $D$ as ${(r, θ) | 0 \leq θ \leq π, 0 \leq r \leq 1 + cos θ}$ as shown in the following figure.

A region D is given as the top half of a cardioid with equation r = 1 + cos theta. — The region $D$ is the top half of a cardioid.

Hence, we have

\begin{matrix} \iint_{D} r^{2} sin θ r d r d θ & = \int_{θ = 0}^{θ = π} \int_{r = 0}^{r = 1 + cos θ} (r^{2} sin θ) r d r d θ \\ = \frac{1}{4} {\int_{θ = 0}^{θ = π} [r^{4}]}_{r = 0}^{r = 1 + cos θ} sin θ d θ \\ = \frac{1}{4} \int_{θ = 0}^{θ = π} {(1 + cos θ)}^{4} sin θ d θ \\ = - \frac{1}{4} {[\frac{{(1 + cos θ)}^{5}}{5}]}_{0}^{π} = \frac{8}{5} . \end{matrix}

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Evaluate the integral

\iint_{D} r^{2} {sin}^{2} 2 θ r d r d θ where D = {(r, θ) | 0 \leq θ \leq π, 0 \leq r \leq 2 \sqrt{cos 2 θ}} .

$π / 8$

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Polar areas and volumes

As in rectangular coordinates, if a solid $S$ is bounded by the surface $z = f (r, θ),$ as well as by the surfaces $r = a, r = b, θ = α,$ and $θ = β,$ we can find the volume $V$ of $S$ by double integration, as

V = \iint_{R} f (r, θ) r d r d θ = \int_{θ = α}^{θ = β} \int_{r = a}^{r = b} f (r, θ) r d r d θ .

If the base of the solid can be described as $D = {(r, θ) | α \leq θ \leq β, h_{1} (θ) \leq r \leq h_{2} (θ)},$ then the double integral for the volume becomes

V = \iint_{D} f (r, θ) r d r d θ = \int_{θ = α}^{θ = β} \int_{r = h_{1} (θ)}^{r = h_{2} (θ)} f (r, θ) r d r d θ .

We illustrate this idea with some examples.

Finding a volume using a double integral

Find the volume of the solid that lies under the paraboloid $z = 1 - x^{2} - y^{2}$ and above the unit circle on the $x y$ -plane (see the following figure).

The paraboloid z = 1 minus x squared minus y squared is shown, which in this graph looks like a sheet with the middle gently puffed up and the corners anchored. — Finding the volume of a solid under a paraboloid and above the unit circle.

By the method of double integration, we can see that the volume is the iterated integral of the form $\iint_{R} (1 - x^{2} - y^{2}) d A$ where $R = {(r, θ) | 0 \leq r \leq 1, 0 \leq θ \leq 2 π} .$

This integration was shown before in [link] , so the volume is $\frac{π}{2}$ cubic units.

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Finding a volume using double integration

Find the volume of the solid that lies under the paraboloid $z = 4 - x^{2} - y^{2}$ and above the disk ${(x - 1)}^{2} + y^{2} = 1$ on the $x y$ -plane. See the paraboloid in [link] intersecting the cylinder ${(x - 1)}^{2} + y^{2} = 1$ above the $x y$ -plane.

A paraboloid with equation z = 4 minus x squared minus y squared is intersected by a cylinder with equation (x minus 1) squared + y squared = 1. — Finding the volume of a solid with a paraboloid cap and a circular base.

First change the disk ${(x - 1)}^{2} + y^{2} = 1$ to polar coordinates. Expanding the square term, we have $x^{2} - 2 x + 1 + y^{2} = 1 .$ Then simplify to get $x^{2} + y^{2} = 2 x,$ which in polar coordinates becomes $r^{2} = 2 r cos θ$ and then either $r = 0$ or $r = 2 cos θ .$ Similarly, the equation of the paraboloid changes to $z = 4 - r^{2} .$ Therefore we can describe the disk ${(x - 1)}^{2} + y^{2} = 1$ on the $x y$ -plane as the region

D = {(r, θ) | 0 \leq θ \leq π, 0 \leq r \leq 2 cos θ} .

Hence the volume of the solid bounded above by the paraboloid $z = 4 - x^{2} - y^{2}$ and below by $r = 2 cos θ$ is

\begin{matrix} V & = \iint_{D} f (r, θ) r d r d θ = \int_{θ = 0}^{θ = π} \int_{r = 0}^{r = 2 cos θ} (4 - r^{2}) r d r d θ \\ = \int_{θ = 0}^{θ = π} [4 \frac{r^{2}}{2} - {\frac{r^{4}}{4} |}_{0}^{2 cos θ}] d θ \\ = \int_{0}^{π} [8 {cos}^{2} θ - 4 {cos}^{2} θ] d θ = {[\frac{5}{2} θ + \frac{5}{2} sin θ cos θ - sin θ {cos}^{3} θ]}_{0}^{π} = \frac{5}{2} π . \end{matrix}

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Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if $f$ has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.

Finding a volume using a double integral

Find the volume of the region that lies under the paraboloid $z = x^{2} + y^{2}$ and above the triangle enclosed by the lines $y = x, x = 0,$ and $x + y = 2$ in the $x y$ -plane ( [link] ).

First examine the region over which we need to set up the double integral and the accompanying paraboloid.

This figure consists of three figures. The first is simply a paraboloid that opens up. The second shows the region D bounded by x = 0, y = x, and x + y = 2 with a vertical double-sided arrow within the region. The second shows the same region but in polar coordinates, so the lines bounding D are theta = pi/2, r = 2/(cos theta + sin theta), and theta = pi/4, with a double-sided arrow that has one side pointed at the origin. — Finding the volume of a solid under a paraboloid and above a given triangle.

The region $D$ is ${(x, y) | 0 \leq x \leq 1, x \leq y \leq 2 - x} .$ Converting the lines $y = x, x = 0,$ and $x + y = 2$ in the $x y$ -plane to functions of $r$ and $θ,$ we have $θ = π / 4,$ $θ = π / 2,$ and $r = 2 / (cos θ + sin θ),$ respectively. Graphing the region on the $x y$ -plane, we see that it looks like $D = {(r, θ) | π / 4 \leq θ \leq π / 2, 0 \leq r \leq 2 / (cos θ + sin θ)} .$ Now converting the equation of the surface gives $z = x^{2} + y^{2} = r^{2} .$ Therefore, the volume of the solid is given by the double integral

\begin{matrix} V & = \iint_{D} f (r, θ) r d r d θ = \int_{θ = π / 4}^{θ = π / 2} \int_{r = 0}^{r = 2 / (cos θ + sin θ)} r^{2} r d r d θ = {\int_{π / 4}^{π / 2} [\frac{r^{4}}{4}]}_{0}^{2 / (cos θ + sin θ)} d θ \\ = \frac{1}{4} {\int_{π / 4}^{π / 2} (\frac{2}{cos θ + sin θ})}^{4} d θ = \frac{16}{4} {\int_{π / 4}^{π / 2} (\frac{1}{cos θ + sin θ})}^{4} d θ = 4 {\int_{π / 4}^{π / 2} (\frac{1}{cos θ + sin θ})}^{4} d θ . \end{matrix}

As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as

V = \int_{0}^{1} \int_{x}^{2 - x} (x^{2} + y^{2}) d y d x .

Evaluating gives

\begin{matrix} V & = \int_{0}^{1} \int_{x}^{2 - x} (x^{2} + y^{2}) d y d x = {\int_{0}^{1} [x^{2} y + \frac{y^{3}}{3}] |}_{x}^{2 - x} d x \\ = \int_{0}^{1} \frac{8}{3} - 4 x + 4 x^{2} - \frac{8 x^{3}}{3} d x \\ = {[\frac{8 x}{3} - 2 x^{2} + \frac{4 x^{3}}{3} - \frac{2 x^{4}}{3}] |}_{0}^{1} = \frac{4}{3} . \end{matrix}

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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