# Triple Integral in Cartesian Coordinates

1. Nov 9, 2008

### daveyman

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

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2. Nov 9, 2008

### 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?

3. Nov 9, 2008

### 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?

4. Nov 9, 2008

### daveyman

Post Deleted.

Last edited: Nov 9, 2008
5. Nov 9, 2008

### gabbagabbahey

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?

6. Nov 9, 2008

### daveyman

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

7. Nov 9, 2008

### 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?

8. Nov 9, 2008

### daveyman

Post Deleted.

Last edited: Nov 9, 2008
9. Nov 9, 2008

### daveyman

Post Deleted.

10. Nov 9, 2008

### 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!

11. Nov 9, 2008

No problem