Double Integral with Polar Coordinates

[cse63146]

Homework Statement

\int^{0}_{-3}\int^{\sqrt{9 - x^2}}_{- \sqrt{9 - x^2}} \sqrt{1 + x^2 + y^2} dy dx

Homework Equations

x = rcos(theta)
y = rsin(theta)

The Attempt at a Solution

By making \sqrt{9 - x^2} = y then changing it to polar coordinates, I got r to be +/-3

but I'm not sure how to find what theta is bounded by. From what I read, since its \sqrt{1 + x^2 + y^2} it's in the first quadrant. Not sure what to do now.

 

[Mark44]

Your region of integration is the left half of a circle of radius 3, centered at the origin. I would let r run from 0 to 3, theta from pi/2 to 3pi/2.

 

[cse63146]

How did you know that theta runs from pi/2 to 3pi/2?

 

[HallsofIvy]

How did you know that theta runs from pi/2 to 3pi/2?

He just told you:

Your region of integration is the left half of a circle of radius 3, centered at the origin.

\theta= 0 is in the direction of the positive x-axis, \theta= \pi/2 points in the direction of the positive y-axis, \theta= \pi points in the direction of the negative x-axis, and \theta= 3\pi/2 points in the direction of the negative x-axis. The cover the left half of the circle, you go counterclockwise from the positive y-axis to the negative x-axis: from \pi/2 to 3\pi/2.

 

[cse63146]

The cover the left half of the circle, you go counterclockwise from the positive y-axis to the negative x-axis: from \pi/2 to 3\pi/2.

Can't believe I didn't realize that; I'm such an idiot. Thank you both.

I'm probably wrong, but wouldn't \theta= 3\pi/2 point to the negative y - axis, since it's the left half of a circle?

 

[Mark44]

Right. I'm sure that's what Halls meant.

 

[cse63146]

So I tried another one, but the book says I'm wrong:


\int^{1}_{0}\int^{x}_{0} \frac{x + y}{\sqrt{x^2 + y^2}} dx dy = \int^{\pi / 2}_{0}\int^{1}_{0} \frac{rcos\vartheta + rsin \vartheta}{\sqrt{r^2 cos^2\vartheta + r^2 sin^2 \vartheta}} r dr d \vartheta = \int^{\pi / 2}_{0}\int^{1}_{0} \frac{rcos\vartheta + rsin \vartheta}{r} r dr d \vartheta = 1/2 \int^{\pi / 2}_{0} (sin\vartheta - cos \vartheta ) r^2 |^1_0 d \vartheta =

1/2 \int^{\pi / 2}_{0} sin \vartheta - cos \vartheta d\vartheta

when I change the boundry for x to polar coordinates, would it go from 0 - 1 or 0 - rcos?

 

[Mark44]

Your limits on the the first two iterated integrals are wrong. One of the key points in switching from Cartesian to polar is understanding exactly what the region is over which integration takes place.

For the first integral, it looks like you copied the order of integration incorrectly, and it should by dy, then dx, not the other way around as you have them.

Can you describe the region on which integration takes place in the first integral? It's a very simple geometric object. When you have that figured out, you should be able to get the right limits on your polar integral.

 

[cse63146]

For the first integral, it looks like you copied the order of integration incorrectly, and it should by dy, then dx, not the other way around as you have them.

It's the way in my book, unless I have to change them.

For the first integral, y is bounded between 0 - 1, while x is bounded on the positive x-axis (since it's 0 - x).

dx dy = r dr d(theta) or is it dy dx = r dr d (theta)?

 

[Mark44]

It's the way in my book, unless I have to change them.

For the first integral, y is bounded between 0 - 1, while x is bounded on the positive x-axis (since it's 0 - x).

That's a typo in your book. If it were as you are interpreting it, the region would be infinitely large. x is bounded by what, x?
That's the same as saying that the region for the inner integral is from x = 0 to x = x, and for the outer integral, y = 0 to y = 1.

I'm sure this is what they meant:
\int_{x = 0}^1 \int_{y = 0}^x <your f(x, y)> dy dx
and got the differentials in the wrong order.

dx dy = r dr d(theta) or is it dy dx = r dr d (theta)?

Yes.
As far as areas are concerned dx*dy = dy*dx, and they both equal r*dr*d(theta). The difference is that the order of dx and dy controls the order in which integration is done.

 

[cse63146]

so y is bounded between 0 and x, while x is bounded by 0 and 1. That does make more sense. It looks like it's in the first quadrant. So changing it to polar coodrinates:

\int_{0}^{\pi / 4} \int_{0}^{cos\vartheta} r dr d \vartheta

 

[Mark44]

The limits look good, so I trust that you understand the integration region. Don't forget your integrand, though. The integral you have just computes the area of the region, and that's not what you want.

 

[cse63146]

Thanks for your help. Just got one more question. I'm wondering if how I changed the boundries from rectangular coordiantes to polar coordinates is the correct way, and I didnt get the right answer by some fluke.

in \int_{0}^{\pi / 4} \int_{0}^{cos\vartheta} r dr d \vartheta

I did this: y = x => rsin = rcos => sin = cos (and this can only be true when the angle is (pi/4)

for \int^{0}_{-3}\int^{\sqrt{9 - x^2}}_{- \sqrt{9 - x^2}} \sqrt{1 + x^2 + y^2} dy dx

I let y = \sqrt{9 - x^2}\Rightarrow rsin = \sqrt{9 - r^2 cos^2} which eventually gave me r = +/- 3.

Sorry for being so annoying.

 

[Mark44]

On thinking about it some more, your upper limit of integration needs to be r = 1/cos(theta). The limits of integration x = 0 to x = 1 and y = 0 to y = x represent a right triangle with vertices at (0,0), (1, 0) and (1, 1). It's isosceles with angles of 45 degrees, for which sin(theta) = cos(theta) or tan(theta) = 1. In the polar integration r ranges from 0 to the line x = 1, or rcos(theta) = 1, or r = 1/cos(theta).

For the other (actually the first integral in this thread), the region is the left half of a circle whose radius is 3 and where theta ranges between pi/2 and 3pi/2 (already discussed).

In converting from one type of iterated integral to another, the limits of integration depend completely on the region, so if you don't understand what the region looks like, you won't get them right in the new integral.

 

[cse63146]

but I thought r ranges from 0 to the line y=x (or is the same as y = x = 1?)

 

[Knissp]

If r varied from 0 to 1 and theta varied from 0 to pi/4, you would have a region that is the sector of a circle. It would have the curvature because r would include all values between 0 and 1 on this eighth of a circle.

Your region, on the other hand, is a right triangle. Do you see the difference?

 

[Mark44]

but I thought r ranges from 0 to the line y=x (or is the same as y = x = 1?)

Not the same. y = x = 1 is the point (1, 1).