### Examples of Double Integrals in Polar Coordinates (Example 3)

We want to find the value of the following definite integral by converting the double integral into polar coordinates.

$\int_0^a {\int\limits_{ - \sqrt {{a^2} - {y^2}} }^0 {\left( {{x^2}y} \right)dxdy} }$

Here, "a" represents some unknown constant. It'll be a little trickier than normal questions because you'll have to keep carrying that "a" term around, but it's really good practice.

First step is to visualize the region of integration, which is just sketching the boundaries on the integrals as if they were equations you could plot. Here we have "dxdy", so the first integral boundaries (I'm reading from left to right) are for the y variable, while the second integral boundaries are for the x variable.

Looking at $$- \sqrt {{a^2} - {y^2}}$$, it seems a bit tough to visualize. A general equation for a circle is $${x^2} + {y^2} = {a^2}$$, where "a" would represent the radius of the circle (you square root the $${a^2}$$ to get "a"). So if you rearranged the equation for a circle in term of x, it'd look like what we have.

The best way to visualize this though is to just figure out all of the boundaries. The easiest is the first integral boundaries, y is from 0 to "a". So we know "a" must be positive. As well, x goes from $$- \sqrt {{a^2} - {y^2}}$$ to 0, so basically x goes from negative to 0. The only area on a plot where this makes sense is the second quadrant (the top-left quadrant), so it's a quarter of a circle, basically a circle only existing in the second quadrant.

So we know our new polar coordinate boundaries from this. Radius is from 0 to "a". And since the circle only is in the second quadrant, theta is from $$\frac{\pi }{2}$$ to $$\pi$$. And we must not forget to multiply by an extra "r" term when converting from Cartesian to polar coordinates. Substituting all that into the original integral,

$\int_{\frac{\pi }{2}}^\pi {\int\limits_0^a {{r^2}{{\cos }^2}\left( \theta \right)r\sin \left( \theta \right)rdrd\theta } }$

$\int_{\frac{\pi }{2}}^\pi {\int\limits_0^a {{r^4}{{\cos }^2}\left( \theta \right)\sin \left( \theta \right)drd\theta } }$

$\int_{\frac{\pi }{2}}^\pi {\frac{1}{5}} {a^5}{\cos ^2}\left( \theta \right)\sin \left( \theta \right)drd\theta$

$\frac{{{a^5}}}{5}\int_{\frac{\pi }{2}}^\pi {\sin \left( \theta \right){{\cos }^2}\left( \theta \right)d\theta }$

Remember, when there's a mixture of sines and cosines, just use substitution. There's no wrong way because sine and cosine are derivatives of each other. We must also change the integral boundaries when substitution is used, shown below.

$\begin{array}{l}u = \cos \left( \theta \right)\\du = - \sin \left( \theta \right)d\theta \\u\left( {\frac{\pi }{2}} \right) = \cos \left( {\frac{\pi }{2}} \right) = 0\\u(\pi ) = \cos (\pi ) = - 1\\\frac{{{a^5}}}{5}\int_0^{ - 1} { - {u^2}du} \\\left( {\frac{{{a^5}}}{5}} \right)\left( {\frac{{ - 1}}{3}} \right)\left[ {{{\left( { - 1} \right)}^3} - {{(0)}^3}} \right]\end{array}$

Then we simplify a little more and obtain the final answer.

$= \frac{{{a^5}}}{{15}}$

It's not much more difficult, the only tricky part was visualizing what the region of integration looked like and carrying the "a" throughout. Since "a" is not a function of r or theta, it's just a constant, and constants can be immediately taken out of the integral.

Please see other examples of solving double integrals in polar coordinates below:

Double Integrals in Polar Coordinates (Ex1)

Double Integrals in Polar Coordinates (Ex2)

Double Integrals in Polar Coordinates (Ex4 Cylinders and Ellipsoids)

Double Integrals in Polar Coordinates (Ex5 Spheres and Cylinders)

Alternatively, return to the solving double integrals using polar coordinates hub page:

Double Integrals in Polar Coordinates