Finding the bounds of a triple integral in cylindrical coordinates?

[Lauren72]

Homework Statement

I took a picture of the problem so it would be easier to understand.

All I need to know is what the bounds are.

Homework Equations

In cylindrical:

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

The Attempt at a Solution

I don't know why we should change this to cylindrical. I feel like it's easier in cartesian. I solved it the normal way and easily got -160/3. But, if we must do cylindrical I already know what the function is. It's just r^2cos^2(Theta)z. What I cannot figure out is the bounds. I'm really confused because of the y bounds. When I try to draw the image I get the parabola formed by the first bound, but I can't figure out how it intersects the rest of the limits, particularly y=x. How do you find those bounds? What are they?

Please help! I'm studying for my Calc III final on Monday.

 

[HallsofIvy]

It may well be easier to evaluate in Cartesian coordinates, but that is not the point of the problem! The point is precisely to practice changing from one coordinate system to another.

I recommend drawing the given limits. The "outer" integral, with respect to x, has lower limit 0 and upper limit 2. Okay, draw xy-axes, draw vertical lines at x= 0 (the y-axis) and at x= 2. That marks the "left" and "right" bounds of the area. The "middle" integral, with respect to y, has limits of 0 and x. Draw the horizontal line y= 0 (the x-axis) and the line y= x. You should see that that forms a triangle with vertices at (0, 0), (2, 0), and (2, 2). The "outer" integral, with respect to z, has limits 0 and 4- x^2 so the lower boundary is the xy-plane and the upper boundary is the "parabolic cylinder" given by z= 4- x^2 (since there is no "y" in the formula, the parabola extends up the y- direction).

However, since you only want to change to cylindrical coordinates, you don't need to worry changing that, except to note that x= r cos(\theta) so 4- x^2= 4- r^2 cos^2(\theta). The "inner" integral, with respect to z, will be from 0 to 4- r^2 cos^2(\theta).

Now look at that triangle. Isn't it easy to see that the angle, \theta ranges from x-axis to the line y= x? What angles are those? And, for each angle, r ranges from the origin (r= 0, of course) to the line x= 2. Since x= r cos(\theta), that is the same as r= 2/cos(\theta)= 2 sec(\theta).

 

[Lauren72]

Okay. So Theta ranges from 0 to y=x, therefore it goes from 0 to pi/4, yes?

So when you're finding the bounds for r and theta, is it necessary to draw the 3d graph? It looks like you can find the bounds just by drawing it on the xy plane.

And since we're in cyclindrical, and not spherical, that's why we don't have to worry about doing anything strange to the z bounds, right? We just do a direct substitution?