Converting a Cartesian Integral to a Polar Integral

[Amadeo]

Homework Statement

Convert the following double integral to polar coordinated and evaluate:

Relevant Equations

x=rcosΘ
y=rsinΘ

the graph of x= √4-y^2 is a semicircle or radius 2 encompassing the right half of the xy plane (containing points (0,2); (2,0); (0-2))
the graph of x=y is a straight line of slope 1

The intersection of these two graphs is (√2,√2)

y ranges from √2 to 2. Therefore, the area over which we integrate is between the line y=√2; y=2; and x=√4-y^2 as in the shaded region in this picture:

This means that the new integral, in polar form is:

∫∫ r dr dΘ with r ranging from √2cscΘ to 2 and Θ ranging from π/4 to π/2
However, the solution is

∫∫ r dr dΘ with r ranging from 2cscΘ to 2 and Θ ranging from π/4 to π/2

I do not understand why this is the case. If y=√2, then rsinΘ = √2 => r=(√2)cscΘ. this would then be the lower limit, as r extends out to the upper limit of 2. Thank you for your assistance.

 

[tnich]

From your drawing, it looks to me as though you are letting $x$ vary between $0$ and $\sqrt {4-y^2}$.

 

[Amadeo]

Oh gosh, how embarrassing. Thank you.

 

[HallsofIvy]

The first thing I notice is that (\sqrt{2},\sqrt{2}) lies on the line \theta= \frac{\pi}{4} and that (0, 2) lies on the line \theta= \frac{\pi}{2}. So the "\theta" integral is from \frac{\pi}{4} to \frac{\pi}{2}. Now, for each \theta, r is measured on the line through the origin making angle \theta with the x-axis so with slope tan(\theta): y= tan(\theta)x. That line crosses y= \sqrt{2} where x= \frac{\sqrt{2}}{tan(\theta)}. r is the distance from the origin to that point: r= \sqrt{2+ 2 tan^2(\theta)}= \sqrt{2}\sqrt{1+ tan^2(\theta)}= sec(\theta). That line crosses the circle x^2+ y^2= 4 where x^2+ tan^2(\theta)x^2= sec^2(\theta)x^2= 4 or x= 2 cos(\theta), y= 2 cos(\theta)tan(\theta)= 2 sin(\theta). <br /> <br /> The integral is \int_{\pi/4}^{\pi/2} \int_{sec(\theta)}^{2 sin(\theta)} r dr d\theta.