### Examples of Double Integrals in Polar Coordinates (Example 1)

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

$\int_{ - 3}^3 {} \int_0^{\sqrt {9 - {x^2}} } {\sin ({x^2} + {y^2})dydx}$

The first thing to do is change the coordinates to polar coordinates and determine new lower and upper integral bounds. In general, use the substitutions listed below:

$\begin{array}{l}x = r\cos (\theta )\\y = r\sin (\theta )\\{r^2} = {x^2} + {y^2}\\dxdy = rdrd\theta \end{array}$

Also, you need the integral bounds, which will be in terms of r and theta. The best thing to do is to do a little sketch of what the plot looks like in the x and y coordinates. Notice that x is from -3 to 3 in that integral, you know quadrants 1 and 2 have something in the plot. But the y integral goes from 0 to $$\sqrt {9 - {x^2}}$$, which means nothing exists below the y axis. The upper bound on y is that equation, which is just a circle of radius 3 (you square root the 9) Basically, a sketch of the region of integration in x and y is just a hemi-circle existing only above the y-axis.

We need to take these x and y coordinates and somehow get new integral boundaries. We know x and y coordinate region of integration is over a hemi-circle over the y axis. Remember that we have $$\sqrt {9 - {x^2}}$$, which looks like an equation for a circle. All you need to do is square root the 9 to get 3. So basically this hemicircle has a maximum radius of 3. And the hemicircle starts at radius 0 as well. So the bounds on r is from 0 to 3. Easy! For theta, you need to see how far the circle you sketched goes. It's a hemicircle above the y-axis, and basically for each quadrant you go through you need to add $$\frac{\pi }{2}$$. This hemicircle exists above the y-axis, so it's just going to the second quadrant, so theta goes from 0 to $$\pi$$. Remember, you start at the positive x axis, and you move counter-clockwise.

We also need to substitute $${r^2} = {x^2} + {y^2}$$ into the argument of the sine, and also add an extra "r" when we transform to polar coordinates. Substituting all of this into the original integral, we get:

$\int_0^\pi {} \int_0^3 {\sin ({r^2})rdrd\theta }$

To solve this, we need to use regular substitution. Let $$u = {r^2}$$, then $$du = 2rdr$$. Or, $$rdr = \frac{{du}}{2}$$. Doing this changes the integral boundaries as well, so we need to square the r boundaries. Substituting these in, $\int_0^\pi {} \int_{{0^2}}^{{3^2}} {\frac{1}{2}\sin (u)dud\theta }$

And we solve the inner integral we get $\int_0^\pi {\frac{1}{2}} \left( {1 - \cos (9)} \right)d\theta$

You can simplify sines and cosines if you have a term with $$\pi$$ in them, but cos(9) can't really be simplified. Just work through the final integral again, which is just multiplying it by theta and evaluating it at both integral boundaries. You will end up with the following as your final expression.

$\frac{{\pi \left( {1 - \cos (9)} \right)}}{2}$

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

Double Integrals in Polar Coordinates (Ex2)

Double Integrals in Polar Coordinates (Ex3)

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