In rectilinear or Cartesian coordinates, we locate a point in the plane by specifying how far to travel right or left along the $x$-axis to be just below (or above) the point, then how far to travel up or down to find its 2D location. We specify that coordinate as $(x, y)$.
Cartesian coordinates are great for specifying linear or rectangular things. For example, the line $x = 1$ is the vertical line passing through $x = 1$, for which the $y$-coordinate can be anything at all. Sometimes we say that lines are "native" the Cartesian coordinate system.
A circle has a somewhat more complicated expression in Cartesian coordinates. For example, the unit circle centered at the origin has the equation
$$x^2 + y^2 = 1$$
Now check out polar coordinates.
In the polar coordinate system we represent each point by a distance from the origin, which we call $r$ — we think of it like a vector — and an angle through which we rotate that vector counterclockwise from 0˚, as shown. A coordinate in the polar system has the form $(r, \theta)$.
To express a line in polar coordinates involves using sines and cosines — a bit cumbersome. Lines are not "native" to polar coordinates. On the other hand, the equation of a circle of radius $r = 1$ is just
$$r = 1$$
That is $\theta$ can be anything. Any given function might be more easily dealt with in polar coordinates than Cartesians. We'll explore some of those in this section.
The relationships between polar coordinates $(r, \; \theta)$ and Cartesian coordinates $(x,\; y)$ are:
$$ \begin{align} x &= r \, cos(\theta) \\[5pt] y &= r \, sin(\theta) \end{align}$$
$$ \begin{align} r^2 &= x^2 + y^2 \\[5pt] \theta &= tan^{-1}\left( \frac{y}{x} \right) \end{align}$$
When we used double integrals to find the volume under a function like $f(x, y)$ over an area $A$, we divided that area into squares of dimensions $\Delta A = \Delta x \times \Delta y$.
By taking the limit as $\Delta x \rightarrow 0$ and $\Delta y \rightarrow 0$, we arrive at our differential elements for integration, $dx \, dy$.
We'll need an area element in order to integrate polar functions, $f(r, \theta)$ over some area $A$. In the next figure notice that we define such an area, $\Delta A$, by moving a distance $r$ from the origin and extending this by a small amount $\Delta r$, and rotating the $r$ vector through an angle $\theta$, extended by a small amount $\Delta \theta$.
Now to find the area of $\Delta A$ we have to be a little more careful. At first we might think it's $\Delta r \Delta \theta$, but notice that the units here (length × angle) is not a unit of area. We recall that the length of the arc along the outside of $\Delta A$ is $r \, \Delta \theta$. Thus the area of $\Delta A$ is $\Delta A = r \Delta \theta \, \Delta r$.
If we take the limits as $\Delta r \rightarrow 0$ and $\Delta \theta \rightarrow 0$, we find
$$dA = r \, dr \, d\theta.$$
So we'll have to remember that a double integral over an area $A$ in polar coordinates will have the form
$$\int \int_A f(r, \theta) \, r \, dr \, d\theta$$
Let's calculate the volume of a sphere of radius $r = 1$ using double integrals. We already know this volume, of course: $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \text{ units}^3$, so we'll know the answer in advance when we try these other methods. We'll first do this in Cartesian coordinates, then we'll convert to polar coordinates and see how much simpler this kind of double integral becomes.
Here's our sphere, centered at the origin. Its equation is
$$x^2 + y^2 + z^2 = 1,$$
which we can write in function form as $z = f(x,y) = \pm \sqrt{1-x^2-y^2}$.
We can focus just on the upper hemisphere, for which
$$z = f(x,y) = \pm \sqrt{1-x^2-y^2}$$
The region over which we're integrating would be the circle $x^2 + y^2 = 1$ in the $xy$-plane.
The
We can do even better in terms of symmetry, however, by cutting the hemisphere in half and multiplying its volume by 8.
The double integral is then
$$V = 8 \int_0^1 dx \int_0^{\sqrt{1-x^2}} dy \, (1-x^2-y^2)$$
We'll need a couple of tricks to do this integral. The first is, notice that the first integration is along the $y$-coordinate, then we consider $1-x^2$ to be a constant. Let's substitute $a^2$ for that:
$$V = 8 \int_0^1 dx \int_0^a dy (a^2 - y^2)^{\frac{1}{2}}$$.
Now we can use a trigonometric substitution: Let $y = a \, sin(t)$. Then we can establish these other quantities to help us move forward:
$$ \begin{align} y &= a \, sin(t) \\[5pt] a^2 - y^2 &= a^2 - a^2 sin^2(t) \\[5pt] &= a^2 cos^2(t) \\[5pt] dy &= a \, cos^2 (t) \, dt \end{align}$$
$$ \begin{align} \text{when } y = 0, \; t = 0 \\[5pt] \text{when } y = a, \; t = \frac{\pi}{2} \end{align}$$
Our integral now converts to
$$ \begin{align} 8 \int_0^1 dx &\int_0^{\frac{\pi}{2}} [a^2 \, cos^2(t)]^{\frac{1}{2}} a \, cos(t) \, dt\\[5pt] &= 8 \int_0^1 dx \int_0^{\frac{\pi}{2}} dt [a^2 cos^2(t)] \end{align}$$
Now we can use the power reduction formula, $cos^2(t) = \frac{1}{2} (1+cos(2t)) t$ to eliminate the $cos^2$ term:
$$ \begin{align} V &= \frac{8}{2} \int_0^1 dx \int_0^{\frac{\pi}{2}} a^2 dt [1 + cos(2t)] \\[5pt] &= 4 \int_0^1 dx \left[t + \frac{1}{2} sin(2t) \right]_0^{\frac{\pi}{2}} \end{align}$$
Now we'll reinstate $a^2 = 1-x^2$ as we do the $dx$ integral:
$$ \require{cancel} \begin{align} V &= 4 \int_0^1 (1-x^2)[t + sin(2t)]_0^{\frac{\pi}{2}} \\[5pt] &= \frac{4 \pi}{2} \int_0^1 dx \, (1-x^2) \\[5pt] &= \frac{4 \pi}{2} \left(x - \frac{x^3}{3} \right) \bigg|_0^1 \\[5pt] &= \frac{4 \pi}{2} \left( 1 - \frac{1}{3} - 0 - 0 \right) \\[5pt] &= \frac{4 \pi}{\cancel{2}} \cdot \frac{\cancel{2}}{3} \\[5pt] &= \frac{4 \pi}{3} \text{ units}^3 \end{align}$$
Just what we expected.
Here's our setup in polar coordinates. The integration along $\theta$ will go from $0$ to $\frac{\pi}{2}$ and along $r$ from $0$ to $1$.
The equation of our sphere is
$$z = \sqrt{1-(x^2+y^2)}$$
In polar coordinates, we know that $x^2 + y^2 = r^2$, so our function in polar coordinates is
$$f(r, \theta) = \sqrt{1-r^2}$$
The double integral is then
$$V = 8 \int_0^{\pi/2} d \theta \int_0^1 (\sqrt{1-r^2}) r \, dr$$
We can do this integral with a simple u-substitution like this:
$$ \begin{align} \text{Let } \; u &= 1-r^2 \\[5pt] \text{then } \; du &= -2 r \, dr \end{align}$$
$$ \begin{align} r = 0 \; \rightarrow \; u = 1 \\[5pt] r = 1 \; \rightarrow \; u = 0 \end{align}$$
$$ \begin{align} V &= -\frac{8}{2} \int_0^{\pi/2} d\theta \int_1^0 u^{\frac{1}{2}} \, du \\[5pt] &= -4 \int_0^{\pi/2} d \theta \; \frac{2}{3} u^{\frac{3}{2}} \bigg|_1^0 \\[5pt] &= - \frac{8}{3} \int_0^{\pi/2} d \theta \; u^{\frac{3}{2}} \bigg|_1^0 \\[5pt] &= - \frac{8}{3} \int_0^{\pi/2} d \theta \; [0-1] \\[5pt] &= \frac{8}{3} \int_0^{\pi/2} d \theta \\[5pt] &= \frac{8}{3} \theta \bigg|_0^{\pi/2} = \frac{8}{3} \pi \cdot \frac{\pi}{2} \\[5pt] &= \frac{4 \pi}{3} \text{ units}^3 \end{align}$$
Much simpler, right? One of the tricks to finding such volumes is to determine which coordinate system will simplify the problem the most. In this case, polar coordinates are clearly superior.
Find the area of one loop of the "four-leafed rose," $r = cos(2 \theta)$.
Here is a polar plot of this function. We're looking for the shaded area.
We're looking for the double integral
$$A(R) = \int \int_R dA = \int_{-\pi/4}^{\pi/4} d \theta \int_0^{cos(2\theta)} r \, dr$$
The
The
The solution is
$$ \begin{align} A(R) &= \int_{-\pi/4}^{\pi/4} d \theta \int_0^{cos(2\theta)} r \, dr \\[5pt] &= \int_{-\pi/4}^{\pi/4} d \theta \left[ \frac{1}{2} r^2 \right]_0^{cos(2 \theta)} \\[5pt] &= \frac{1}{4} \int_{-\pi/4}^{\pi/4} d \theta (1+cos(4\theta)) \tag{1}\\[5pt] &= \frac{1}{4} \left[ \theta + \frac{1}{4} sin(4 \theta) \right]_{-\pi/4}^{\pi/4} \\[5pt] &= \frac{\pi}{8} \text{ units}^2 \end{align}$$
In step (1) above we used the power reduction formula for $cos^2(t)$,
$$cos^2(t) = \frac{1+cos(2t)}{2}.$$
Convert the integral $\int \int_R x^2 y \, dA$ to polar coordinates, then integrate it above the half circle of radius $r = 5$ centered at the origin.
The circle of radius $r = 5$ has just that equation in polar coordinates. Our limits of integration will be $0 \rightarrow 5$ along the $r$ coordinate, and $0 \rightarrow \pi$ in the $\theta$ coordinate.
We'll need to convert the function, $f(x,y) = x^2 y$ into polar coordiates using
$$ \begin{align} x &= r \, cos(\theta) \\[5pt] y &= r \, sin(\theta) \end{align}$$
The function converts to $f(r, \theta) = r^2 \, cos^2(\theta) \, r \, sin(\theta)$. Then the double integral is
$$\int_0^{\pi} d\theta \int_0^5 \, r^3 cos^2 (\theta) sin(\theta) r \, dr$$
Here are the steps in solving this integral:
$$ \begin{align} V &= \int_0^{\pi} d \theta \int_0^5 cos^2(\theta) sin(\theta)\; r^4 \, dr \\[5pt] &= \int_0^{\pi} d \theta \left[cos^2(\theta) sin(\theta) \; \frac{r^5}{5}\right]_0^5 \\[5pt] &= 5^4 \int_0^{\pi} d \theta \; cos^2(\theta) sin(\theta) \end{align}$$
We can do this integral by u-substitution: Let $u = cos(\theta)$, then $du = - \, sin(\theta) \, d\theta$, so we have
$$ \begin{align} &= -5^4 \int_1^{-1} \, u^2 \, du \\[5pt] &= -5^4 \; \frac{u^3}{3} \bigg|_1^{-1} \\[5pt] &= -5^4 \cdot \left( -\frac{2}{3} \right) \\[5pt] &= \frac{1250}{3} \text{ units}^3 \end{align}$$
Solve the following double integrals:
Find the volume of the function $f(x,y) = x+y$ over the region $R$, where $R$ is the region in the first quadrant of the $xy$ plane inside the circle $x^2+y^2 \le a^2$ and under the line $y = \sqrt{2} x$.
Here's the region over which we're integrating the function $f(x,y)=x+y$.
To do this integral in polar coordinates, we'll have to find the angle $\theta$ shown in the graph. That's just a matter of a little bit of trigonometry. The
$$\theta = tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{\sqrt{2} x}{x} \right) = \sqrt{2}$$
So we have $\theta = tan^{-1}(\sqrt{2}) = 0.955 \text{ rad}$ (about 54˚). So our limits of integration along $r$ will be $0$ to $a$ and along the $\theta$ coordinate they'll be $0$ to $0.955$. Now we need to convert the function $f(x,y)$ to $f(r,\theta)$. We just do a simple substitution, $x = r \, cos(\theta)$ and $y = r \, sin(\theta)$:
$$f(r, \theta) = r(sin(\theta) + cos(\theta))$$
Now our integral is
$$\int_0^{0.955} d\theta \int_0^a dr \; r^2(sin(\theta) + cos(\theta))$$
Now to solve the integral:
$$ \begin{align} \int_0^{0.955} & d\theta \int_0^a dr \; r^2(sin(\theta) + cos(\theta)) \\[5pt] &= \frac{r^3}{3} \bigg|_0^a \, \int_0^{0.955} d\theta \, (sin(\theta) + cos(\theta)) \\[5pt] &= \frac{a^3}{3} \int_0^{0.955} d\theta \, (sin(\theta) + cos(\theta)) \\[5pt] &= \frac{a^3}{3} \left[ -cos{\theta} + sin(\theta) \right]_0^{0.955} \\[5pt] &= \frac{a^3}{3} \left[ -cos{0.955} + sin(0.955) + cos(0) - sin(0) \right] \\[5pt] &= \frac{a^3}{3} \left[ -0.578 + 0.816 + 1 - 0 \right] \\[5pt] &= \frac{a^3}{3} \cdot 1.238 \\[5pt] &= 0.413 a^3 \end{align}$$
Calculate the volume inside of the sphere with equation $x^2 + y^2 + z^2 = 16$ and outside of the cylinder $x^2+y^2 = 4$.
Here is a picture of the volume we're interested in. It's the volume of the sphere of radius 4 minus what's inside the cylinder of radius 2:
The excluded area is the cylinder $x^2 + y^2 = 4$, which is $r = 2$ in polar coordinates (actually that equation is in cylindrical coordinates, which we'll tackle eventually. So we'll be integrating along the $r$ coordinate from $2$ to $4$, and in the $\theta$ coordinate all the way around the circle, from $0$ to $2 \pi$. The equation for the sphere transforms into polar coordinates using $r^2 = x^2 + y^2$ like this:
$$ \begin{align} x^2 + y^2 + z^2 &= 16 \\[5pt] z^2 &= 16 - x^2 - y^2 \\[5pt] z^2 &= 16 - (x^2 + y^2) \\[5pt] z^2 &= 16 - r^2 \\[5pt] z &= \sqrt{16-r^2} \end{align}$$
Now the integral setup is:
$$V = \int_0^{2 \pi} d \theta \int_2^4 dr \, r \, (16-r^2)^{\frac{1}{2}}$$
We can solve this integral by u-substitution:
$$ \begin{align} \text{Let } \; u &= 16-r^2 \phantom{00} \text{ then } \; du = -2r dr \\[5pt] \text{and } \; r &= 2 \rightarrow u = 12, \; r = 4 \rightarrow u = 0 \end{align}$$
Then our integral reduces to
$$ \begin{align} V &= -\frac{1}{2} \int_0^{2 \pi} d \theta \int_{12}^0 u^{\frac{1}{2}} du \\[5pt] &= -\frac{1}{\cancel{2}}\frac{\cancel{2}}{3} \int_0^{2 \pi} d \theta \; u^{\frac{3}{2}} \bigg|_{12}^0 \\[5pt] &= -\frac{1}{3} \int_0^{2 \pi} d\theta \; \left[0-12^{\frac{3}{2}}\right] \\[5pt] &= \frac{1}{3} \cdot 24 \sqrt{3} \cdot \theta \bigg|_0^{2 \pi} \\[5pt] &= \frac{24 \sqrt{3}}{3} \cdot 2 \pi \\[5pt] &= 16 \pi \sqrt{3} \text{ units}^3 \end{align}$$
Calculate the volume that lies between the hyperboloid (two sheets) $-x^2-y^2+z^2 = 1$ and the plane $z = 2$.
Here's a plot of these two functions. The hyperboloid is of two sheets, and the plane $z = 2$ cuts through the top sheet, defining the volume outlined in yellow.
We need to find the intersection of these two figures. Inserting $z = 2$ into the equation of the hyperboloid gives
$$ \begin{align} -x^2 - y^2 + 2^2 &= 1 \\[5pt] -x^2 - y^2 &= -3 \\[5pt] x^2 + y^2 &= 3 \end{align}$$
This is the equation of a circle with radius $r = \sqrt{3}$, so our limits of integration along the $r$ coordinate will be $0 \rightarrow 3$. With respect to the $\theta$ coordinate, we'll integrate from $0$ to $2 \pi$.
Now the hyperboloid can be expressed in polar coordinates using $r^2 = x^2 + y^2$:
$$ \begin{align} -x^2 - y^2 + z^2 &= 1 \\[5pt] x^2 + y^2 &= 1 - z^2 \\[5pt] r^2 + 1 &= z^2 \\[5pt] z &= \pm \sqrt{r^2 + 1} \end{align}$$
So our volume double integral is
$$V = \int_0^{2\pi} d\theta \int_0^3 (r^2+1)^{\frac{1}{2}} \, r \, dr$$
We can solve this integral by u-substitution:
$$ \begin{align} \text{Let } \; u &= r^2+1 \phantom{00} \text{ then } \; du = 2r dr \\[5pt] \text{and } \; r &= 0 \rightarrow u = 1, \; r = \sqrt{3} \rightarrow u = 4 \end{align}$$
This reduced our integral to
$$ \begin{align} V &= \frac{1}{2} \int_0^{2\pi} d\theta \int_1^4 u^{\frac{1}{2}} \, r \, dr \\[5pt] &= \frac{1}{\cancel{2}} \frac{\cancel{2}}{3} \int_0^{2\pi} d\theta u^{\frac{3}{2}} \bigg|_1^4 \\[5pt] &= \frac{1}{3} \int_0^{2\pi} d\theta \, 4^{\frac{3}{2}} - 1 \\[5pt] &= \frac{7}{3} \int_0^{2\pi} d\theta \\[5pt] &= \frac{14 \pi}{3} \text{ units}^3 \end{align}$$
Convert the integral
$$\int_0^a \, dy \int_{-\sqrt{a^2-x^2}} \, dx \, [x^2y]$$
to polar coordinates, and find and interpret the resulting volume.
First we have to interpret the meaning of the limits of integration in the $x$ and $y$ coordinates and see if we can translate them to polar coordinates. The lower limit of integration in the $dx$ integral is $-\sqrt{a^2-y^2}$, which is the left side of a circle of radius $a$ in the $xy$-plane.
$$x^2+y^2=a^2 \; \rightarrow \; x = -\sqrt{a^2-y^2}$$
In polar coordinates our limits of integration will be $0$ to $a$ in the $r$ coordinate and either $\left[ \frac{\pi}{2}, -\frac{\pi}{2} \right]$ or $\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$ in the $\theta$ coordinate. Conversion of the function $f(x,y) = x^2 y$ to polar coordinates gives
$$f(r, \theta) = r^3 cos^2(\theta) sin(\theta)$$
So our integral in polar coordinates is
$$V = \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \int_0^a dr \; r^4 \, cos^2(\theta) sin(\theta)$$
doing the inner integral yields
$$ \begin{align} V &= \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \int_0^a dr \; r^4 \, cos^2(\theta) sin(\theta) \\[5pt] &= \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \left[ \frac{r^5}{5} \right]_0^a \, cos^2(\theta) sin(\theta) \\[5pt] &= \frac{a^5}{5} \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta cos^2(\theta) sin(\theta) \\[5pt] \end{align}$$
The $\theta$ part can be done by simple u-substitution:
$$ \begin{align} \text{Let } \; u &= cos(\theta), \; \text{ then } \; du = sin(\theta) d\theta \\[5pt] \theta &= \frac{\pi}{2} \; \rightarrow \; u = 0 \\[5pt] \theta &= \frac{3\pi}{2} \; \rightarrow \; u = 0 \end{align}$$
Making the substitutions and changing the limits of integration gives
$$V = \frac{a^5}{5} \int_0^0 du \, u^2$$
OK, now that's a weird result: The integral is zero when the limits are the same, so this volume must be zero. What's going on? Here are a couple of pictures of the function that will help.
Here is the function $f(x,y) = x^2y$. It has two-fold symmetry along the $x$ and $y$-axes around the origin.
If we look at this function from above and we draw our semicircular area of integration over it, we see that the upgoing (+) regions and donwgoing (-) regions are symmetric, so the volume has as much below as above the plane z=0, thus those positive and negative volumes cancel.
Use double integration in polar coordinates to calculate the area enclosed by the two cardioid $r = 1+cos(\theta)$ and $r = 1-cos(\theta)$.
Here are the two cardioids. The area enclosed by both is highlighted:
If we look just at the curve
$$r = 1-cos(\theta),$$
we see that the region graphed between $\theta = -\frac{\pi}{2}$ and $\theta = \frac{\pi}{2}$ is half of the region of interest, so we can do the integral of that region and multiply by $2$.
Our integral and its solution are:
$$ \begin{align} A &= 2 \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} d\theta \int_0^{1-cos(\theta)} \, dr \\[5pt] &= 2 \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} d\theta [r]_0^{1-cos(\theta)} \\[5pt] &= 2 \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} d\theta [1-cos(\theta)] \\[5pt] &= 2 \left[ \theta - sin(\theta) \right]_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\[5pt] &= 2 \left[ \frac{\pi}{2} - 1 + \frac{\pi}{2} - 1 \right] \\[5pt] &= 2 \left[ \frac{\cancel{2} \pi}{\cancel{2}} - 2\right] \\[5pt] &= \frac{\pi}{2}-1 \text{ units}^2 \end{align}$$
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