# Thread: Triple integral

1. ## Triple integral

f(x,y,z) = z

Integrate the region W, which is bounded by

z = 0
x^2 + 4y^2 = 4
z = x + 2

So I think the "bottom" of the region in the xy-plane should form an elipse, and the "top" is just a plane tilted in the x-direction. This means that the region is a type 3-region in the plane, and a type 1 region in space since the top part is bounded by z? I'm struggling to set up the start of this integral problem, so any help would be appreciated!

2. Originally Posted by gralla55
f(x,y,z) = z

Integrate the region W, which is bounded by

z = 0
x^2 + 4y^2 = 4
z = x + 2

So I think the "bottom" of the region in the xy-plane should form an elipse, and the "top" is just a plane tilted in the x-direction. This means that the region is a type 3-region in the plane, and a type 1 region in space since the top part is bounded by z? I'm struggling to set up the start of this integral problem, so any help would be appreciated!
first of all draw yourself figure, that can help a lot in understanding what have to be done, so you can do your problem

3. Well, I don't know if this is correct, but this is how I visualize this problem:

4. Originally Posted by gralla55
Well, I don't know if this is correct, but this is how I visualize this problem:
first of all , your ellipse is with from -2 to 2 o x -axis and from -1 to 1 on y-axis (check your algebra, probably just typo somewhere)

okay now you can do this with or without using change to some coordinates depending how do you wont to solve this you will need to determine limits for either x,y,z or r,theta,phi (all depend on what is more "closer" to you)

5. So for all x and y, z goes from 0 to the plane x+ 1. Those should be the limits on the "dz" integral.

There are several different ways to handle the dxdy integral.

If you choose to integrate with respect to y first and then x, solve for y: $\displaystyle y= \pm\frac{1}{2}\sqrt{4- x^2}$. Those should be the limits on the "dy" integral.

And, of course, x ranges from -2 to +2 as yecKiM said.

6. note that if you have ellipse and you want to use let's say Cylindrical coordinate system, your substitution will be

$\displaystyle x = a\cdot r \cos {\theta}$

$\displaystyle y = b\cdot r \sin {\theta}$

$\displaystyle z=z$

$\displaystyle J = a\cdot b \cdot r$

that's case if you have ellipse with center (0,0)

if your ellipse is shifted on x or y axis , or both ...

$\displaystyle x = a\cdot r \cos {\theta} +p$

$\displaystyle y = b\cdot r \sin {\theta} + q$

and of course J will be different in that case

P.S. limits can be easy extracted form the image (if figure is something that you can and know to draw) or just solving those starting equations like that , or putting substitute one and solve them... either way you will have the same (correct) ...