Problem on a double integral over a circle
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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
Polar coordinates
If a problem involves terms like $x^2+y^2$, it may be convenient to use polar coordinates.
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
Polar coordinates
In polar coordinates, $x = r \cos \theta$ and $y = r \sin \theta$. Thus, $x^2 + y^2 = r^2$.
Double integrals in polar coordinates
In polar coordinates, the differential area element $dx dy = r dr d\theta$.
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
Factoring a double integral
If the bounds of a double integral are constants and the integrand factors, then the integral factors:$$\int_{r=a}^b \int_{\theta=c}^d f(r) g(\theta) d\theta dr = \int_{r=a}^b f(r) dr \int_{\theta=c}^d g(\theta) d\theta$$
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 $$
\iint_D (1 - x^2 -y^2) dxdy = \frac{1}{2} \pi \Bigl(2 a^2 - a^4\Bigr).
$$
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