Question

The integral \(\iint(x^2+y^2)dxdy\) over the area of a disk of radius 2 in the xy plane is ___π.

Answer 1

Correct Answer: 8

To solve the integral \(\iint(x^2+y^2)dxdy\) over the area of a disk of radius 2 in the xy-plane, we use polar coordinates which simplify the integration over circular regions. The transformation is given by \(x = r\cos\theta\) and \(y = r\sin\theta\). The Jacobian of the transformation from Cartesian to polar coordinates is \(r\), thus \(dxdy = rdrd\theta\). The limits for \(r\) will range from 0 to 2, and for \(\theta\) from 0 to \(2\pi\).
The integral becomes:
\(\int_{0}^{2\pi}\int_{0}^{2}(r^2)rdrd\theta=\int_{0}^{2\pi}\int_{0}^{2}r^3drd\theta\).
First, evaluate the inner integral with respect to \(r\):
\(\int_{0}^{2}r^3dr=[\frac{r^4}{4}]_{0}^{2}=\frac{16}{4}=4\).
Next, integrate with respect to \(\theta\):
\(\int_{0}^{2\pi}4d\theta=4[\theta]_{0}^{2\pi}=4(2\pi)=8\pi\).
This result, \(8\pi\), confirms that the integral of \((x^2+y^2)\) over the disk is indeed \(8\pi\).