Polar Coordinates switch bounds double integral?

Question

Switch bounds of this double integral to polar to evaluate

Integral(from 0 to sqrt(2) Integral (from y to sqrt(4-y^2) 1/(1+x^2+y^2) dx dy

Answer 1

Holy jumped up catfish...not even going to touch that. Good luck.

Answer 2

There are three main parts to consider when translating a repeated integral between coordinate systems: (1) region of integration; (2) integrand; (3) differential term (often called the dA term, or dV in triple integration).

Most of the work has to do with the region of integration, which is entirely determined by the limits of integration.

Let's start with the outside integral: y goes from 0 to Sqrt(2). We graph y=0 and y=Sqrt(2) on the same axes. The region of integration will lie between these horizontal lines.

Now we move to the inside integral: x goes from y to Sqrt(4-y^2). We graph x=y and x=Sqrt(4-y^2). The first of these is better known as the line y=x. As for the second, we can rearrange the equation x=Sqrt(4-y^2) into the form x^2+y^2=4, which describes the circle of radius 2, centered at the origin.

We have four bounding curves:
y=0 on the bottom
y=Sqrt(2) on the top
y=x on the left
x^2+y^2=4 on the right

The region is an eighth of a circle, but I prefer to think of it as a slice of pie.

Let's describe the same region in polar coordinates. Good news: this shape was _born_ for polar coordinates.

First, theta goes from 0 (the positive x-axis) to pi/4 (the base angle of our line y=x).

Next, r goes from 0 (the origin) to 2 (the bounding circle).

That's it for the limits of integration; now we should translate the integrand.

The integrand (in rectangular coordinates) is 1/(1+x^2+y^2). Polar and rectangular coordinates are related by these equations:
x=r cos theta
y=r sin theta
x^2+y^2=r^2
tan theta=y/x.
These four equations are always in the background as we translate between polar and rectangular coordinates. Although we might (in principle) use more than one of them, the current problem was built with kindness. We need only the third equation, x^2+y^2=r^2. Using this, we replace the "x^2+y^2" in the original integrand with "r^2", so the integrand becomes:

1/(1+r^2).

Finally, the differential term. In rectangular coordinates, double integrals end with either dx dy or dy dx. In polar coordinates, they usually end with r dr d_theta. Why the extra factor "r" in there? That's another question; I won't answer it here.

To bring it all together, we have theta going from 0 to pi/4 for the outside integral; we have r going from 0 to 2 for the inside integral; our integrand is 1/(1+r^2) with an extra factor of r thrown in because of polar coordinates. That is:

Integral (from 0 to pi/4) Integral (from 0 to 2) r/(1+r^2) dr d_theta

To solve the inside integral, use the substitution u=1+r^2. I'll leave that to you.