## Problem on a double integral over a circle

Let $D$ be the circle of radius $a$ centered at the origin. Evaluate $$\iint_D (1 - x^2 -y^2) dxdy$$
• ## Solution

For convenience, we will let $I = \iint_D (1 - x^2 -y^2) dxdy$.
Recall that
We note that the integrand $1-x^2-y^2$ can be written $1- (x^2 + y^2)$. Hence, we identify the pattern and change to polar coordinates.
Recall that
We can now write the integrand as $$1-x^2 - y^2 = 1 - (x^2 + y^2) = 1 - r^2.$$
Rewriting the double integral in polar coordinates, we get
\begin{align} I = \iint_D (1 - x^2 -y^2) dxdy &= \int_{r=0}^a \int _{\theta=0}^{2 \pi} (1- r^2) r d\theta dr \\ \end{align}
Recall that
Hence, we can write \begin{align}I &= \int_0^a (1-r^2)r dr \int_0^{2\pi} d\theta\\
&= 2\pi \int_0^a (r - r^3) dr\\
&= 2 \pi \Bigl(\frac{1}{2} a^2 - \frac{1}{4} a^4\Bigr)
\end{align}
After simplifying, we get $$\iint_D (1 - x^2 -y^2) dxdy = \frac{1}{2} \pi \Bigl(2 a^2 - a^4\Bigr).$$