Double Integral Problems
Polar Coordinates Double Integrals Problems

Example 1.

Suppose we are given

(1)   \begin{equation*} \int \int_R \frac{y}{x^2+y^2} dA \end{equation*}

where R is the region bounded by portion of the unit disk in first quadrant. We know

(2)   \begin{equation*} x=r \cos \theta,y=r \sin \theta, x^2+y^2=r^2,dA=rdrd\theta \end{equation*}

and since

(3)   \begin{equation*} 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 1 \end{equation*}

we can write

(4)   \begin{equation*} \int _0 ^{\pi /2} \int _0 ^1 \frac{r \cos \theta}{r^2} rdrd\theta = \int _0 ^{\pi /2} \int _0 ^1 \cos \theta dr d\theta =1 \end{equation*}

Example 2.

Suppose we want to evaluate the integral

(5)   \begin{equation*} \int _0 ^2 \int _{\sqrt{2x-x^2}} ^{\sqrt{4x-x^2}}dydx \end{equation*}

We begin by drawing the region

We use polar coordinates as 

(6)   \begin{equation*} x=r \cos \theta,y=r \sin \theta, x^2+y^2=r^2,dA=rdrd\theta \end{equation*}

and so 

(7)   \begin{equation*} \int _0 ^2 \int _{\sqrt{2x-x^2}} ^{\sqrt{4x-x^2}}dydx= \int _{-\pi /2} ^{\pi /2} \int _{0} ^{4 \cos \theta} rdrd\theta - \int _{-\pi /2} ^{\pi /2} \int _{0} ^{2 \cos \theta} rdrd\theta \end{equation*}

Take a close look at this integral and make sure it makes sense. Then compute it.

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