Evaluate iterated integral by converting to polar coordinates.

$\begingroup$

To Find:

$$\int_0^1\int_0^{(1-y^2)^\frac{1}{2}}\cos(x^2+y^2)\,dx\,dy$$

$$=\int_0^\frac{\pi}{2}\int_0^1r\cos(r^2)\,dr\,d\theta$$

$$= \frac{\pi}{2}\sin (1)$$

$\endgroup$ 2

2 Answers

$\begingroup$

The region of integration is the quarter of unit circle in the first quadrant, so passing to polar coordinates we get

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\;,\;\;0\le r\le1\;,\;\;0\le t\le\frac\pi2\implies$$

$$\int_0^1\int_0^{(1-y^2)^{1/2}}\cos(x^2+y^2)dxdy=\int_0^1\int_0^{\pi/2}r\cos r^2dtdr=\frac\pi4\int_0^12r\cos r^2dr=$$

$$=\frac\pi4\left.\sin r^2\right|_0^1=\frac\pi4\sin1$$

I think the difference between us is because there's a tiny mistake there:

$$\int r\cos^2 dr=\frac12\int2r\cos r^2dr=\frac12\sin r^2+K$$

Check this.

$\endgroup$ $\begingroup$

Your answer is correct, except for the small mistake pointed out by Joanpemo. The actual answer is $\frac{\pi}{4\sin(1)}$.

$\endgroup$