Triple Integral in Cartesian Coordinates

[daveyman]

Homework Statement

Use a triple integral to find the volume of the solid enclosed by the paraboloid x=y^2+z^2 and the plane x=16
Note: The triple integral must be performed in Cartesian coordinates.

Homework Equations

The Attempt at a Solution

I calculated the answer numerically using Mathematica (see attached PDF). I've also included a 3D graph to help with visualization. I calculated the answer to be 128\pi, but I have no idea how to set up the integral. I'm guessing the argument of the integral will simply be 1, but I don't know how to construct the bounds.

 

[gabbagabbahey]

The parabaloid opens to the right, so x runs from 0 to 16...Try expressing the z bounds in terms of y and x...for a given x and y, what is the minimum z and maximum z-value in the region?

 

[daveyman]

z=\sqrt{x-y^2}
The maximum value for z would be z=\sqrt{16-y^2}.
The minimum value for z would be 0.
Okay, so I have my x and z bounds, but what do I use for my y bound?

 

[daveyman]

Post Deleted.

 

[gabbagabbahey]

z=\sqrt{x-y^2}
The maximum value for z would be z=\sqrt{16-y^2}
What do I use for y?

Don't you mean z=\pm \sqrt{x-y^2}? ...and you don't want the absolute max and mins for z, you want the max/min for each x,y...which is just z=- \sqrt{x-y^2} for the min, and z=+ \sqrt{x-y^2} for the max...this means that when you integrate over z, you end up with a function of x and y...then you want to integrate over y...what is the boundary of the region in just the x-y plane (ie for z=0)?...what does that mean the max/min of y is for a given value of x?

 

[daveyman]

Good point about the z bounds. So should I use y=\pm\sqrt{16} for my y bounds?

 

[gabbagabbahey]

IF x were always 16, then yes, but for any x, wouldn't you want to use y=\pm \sqrt{x} ? And so when you integrate over y, you get a function of x which you then integrate from 0 to 16...your integral is:

V=\int_{\mathcal{V}} dV= \int_0^{16} \int_{-\sqrt{x}}^{\sqrt{x}} \int_{-\sqrt{x-y^2}}^{\sqrt{x-y^2}} dxdydz

...do you follow?

 

[daveyman]

Post Deleted.

 

[daveyman]

Post Deleted.

 

[daveyman]

\int _0^{16}\int _{-\sqrt{x}}^{\sqrt{x}}\int _{-\sqrt{x-y^2}}^{\sqrt{x-y^2}}1dzdydx = 128\pi

It works beautifully! Thanks again for your help!

 

[gabbagabbahey]

No problem