Answer to Question #185358 in Calculus for Amber

Question

Evaluate the following integral by first converting to an integral in polar coordinates.

X^2dxdy...where x varies from -2 to 0 and y varies from -√(4-y^2) to √(4-y^2)

Answer 1

Given, the integral $\int_{-2} ^{0} \int_{-\sqrt{4-y^2}} ^{\sqrt{4-y^2}} x^2 dxdy$ .

From the integral limit, the curves are

$y=0$ , ..........................................1

$y=-2$, ....................................2

$x^2+y^2=4$ . ..............................3

Convert cartesian to polar coordinate,

$x=rcos\theta,$ ................................4

$y=rsin\theta,$ .................................5

$r=\sqrt{x^2+y^2}$, ........................6

$\theta=tan^{-1}\frac{y}{x}$ ..............................7

From equation 3 and 6,

$r=2$

Therefore, $0\le r \le 2 \ \space and \space 0\le\theta\le2\pi$.

Then, the integral

$\int_{-2} ^{0} \int_{-\sqrt{4-y^2}} ^{\sqrt{4-y^2}} x^2 dxdy \newline =\int_{0} ^{2\pi} \int_{0} ^{2} r^2cos^2\theta(rdr d\theta)\newline=\int_{0} ^{2\pi} \int_{0} ^{2} r^3cos^2\theta dr d\theta\newline =\int_{0} ^{2\pi} [\frac{r^4}{4}]_{0}^{2}\space cos^2\theta dr d\theta\newline =\int_{0} ^{2\pi} 1. cos^2\theta d\theta\newline =\int_{0} ^{2\pi} \frac{1+cos2\theta}{2} d\theta\newline =\frac{1}{2}[\theta+\frac{sin2\theta}{2}]_{0}^{2\pi}\newline =\pi$

Thus, the value of the given integral is $\pi$.