Answer to Question #185358 in Calculus for Amber

Question #185358

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)


1
Expert's answer
2021-05-07T09:09:41-0400

Given, the integral 204y24y2x2dxdy\int_{-2} ^{0} \int_{-\sqrt{4-y^2}} ^{\sqrt{4-y^2}} x^2 dxdy .

From the integral limit, the curves are

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

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

x2+y2=4x^2+y^2=4 . ..............................3

Convert cartesian to polar coordinate,

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

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

r=x2+y2r=\sqrt{x^2+y^2}, ........................6

θ=tan1yx\theta=tan^{-1}\frac{y}{x} ..............................7

From equation 3 and 6,

r=2r=2

Therefore, 0r2  and 0θ2π0\le r \le 2 \ \space and \space 0\le\theta\le2\pi.

Then, the integral

204y24y2x2dxdy=02π02r2cos2θ(rdrdθ)=02π02r3cos2θdrdθ=02π[r44]02 cos2θdrdθ=02π1.cos2θdθ=02π1+cos2θ2dθ=12[θ+sin2θ2]02π=π\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 π\pi.



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