Double Integral in Polar Coordinates: Evaluating and Solving for Limits

[utkarshakash]

Homework Statement

Evaluate the integral by changing into polar coordinates.
\displaystyle \int_0^{4a} \int_{y^2/4a}^y \dfrac{x^2-y^2}{x^2+y^2} dx dy

The Attempt at a Solution

Substituting x=rcos theta and y=rsin theta , the integrand changes to cos 2 \theta r dr d \theta . I know that the region of integration is the area bounded by parabola y^2=4ax and y=x. However, I don't know the limits of r and theta.

 

[Ray Vickson]

Homework Statement

Evaluate the integral by changing into polar coordinates.
\displaystyle \int_0^{4a} \int_{y^2/4a}^y \dfrac{x^2-y^2}{x^2+y^2} dx dy

The Attempt at a Solution

Substituting x=rcos theta and y=rsin theta , the integrand changes to cos 2 \theta r dr d \theta . I know that the region of integration is the area bounded by parabola y^2=4ax and y=x. However, I don't know the limits of r and theta.

Perhaps you should not use $x = r \cos \theta, \: y = r \sin \theta$. Why would you not just do the inner x-integration first, then do the y-integration?

 

[LCKurtz]

Homework Statement

Evaluate the integral by changing into polar coordinates.
\displaystyle \int_0^{4a} \int_{y^2/4a}^y \dfrac{x^2-y^2}{x^2+y^2} dx dy

The Attempt at a Solution

Substituting x=rcos theta and y=rsin theta , the integrand changes to cos 2 \theta r dr d \theta . I know that the region of integration is the area bounded by parabola y^2=4ax and y=x. However, I don't know the limits of r and theta.

Have you drawn a picture of that region? Have you written the equation of that parabola in polar coordinates? You need to do both. Then think of a point inside the region. It will move from $r=0$ to $r$ on the parabola. Once you draw that you should be able to see what values $\theta$ must vary between.

 

[utkarshakash]

Perhaps you should not use $x = r \cos \theta, \: y = r \sin \theta$. Why would you not just do the inner x-integration first, then do the y-integration?

Because I'm required to solve it by changing into polar coordinates.

 

[utkarshakash]

Have you drawn a picture of that region? Have you written the equation of that parabola in polar coordinates? You need to do both. Then think of a point inside the region. It will move from $r=0$ to $r$ on the parabola. Once you draw that you should be able to see what values $\theta$ must vary between.

The polar equation of given parabola will be $r=-2a/(1+cos \theta)$. The limit of theta is pi/4 to pi/2. Integrating the expression wrt r I'm left with $\int_{\pi/4}^{\pi/2} \dfrac{-2a cos 2 \theta}{1+cos \theta} d \theta$

 

[LCKurtz]

The polar equation of given parabola will be $r=-2a/(1+cos \theta)$. The limit of theta is pi/4 to pi/2. Integrating the expression wrt r I'm left with $\int_{\pi/4}^{\pi/2} \dfrac{-2a cos 2 \theta}{1+cos \theta} d \theta$

It looks like you have figured out the limits now, but I don't get the same equation for the parabola. Surely $r$ shouldn't be negative. And let's see the double integral after you have put it in polar coordinates and before you integrate it.