# Volume of the solid enclosed by the paraboloids

1. Oct 28, 2005

### Whatupdoc

Find the volume of the solid enclosed by the paraboloids z= 16(x^2 +y^2) and z=32-16(x^2+y^2)

i'm not sure how i would find the x bounds for this triple integral. here's my work:

16x^2+16y^2 = 32-16x^2+16y^2 => simplifies to y = +- sqrt(1-x^2) (the y-bounds)

z bounds is already given which is from z= 16x^2+16y^2 to 32-16x^2+16y^2

but how would i find the x bounds (mathematically if possible)? i'm not sure

2. Oct 28, 2005

### whozum

I would try a different coordinate system?

3. Oct 28, 2005

### HallsofIvy

Staff Emeritus
Adding the two equations gives 2z= 32 or z= 16. When z= 16 both equations reduce to x^2+ y^2= 1 so you are correct. x ranges from -1 to 1, for each x, y ranges from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$, for each x,y, z ranges from 16(x2+ y[/sup]2[/sup]) to
32- 16(x2+ y2.
The volume is
$$\int_{x= -1}^1 \int{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{z=16(x^2+ y^2)}^{32-16(x^2+ y^2} dzdydx$$
Yes, you would find it simpler to take advantage of the circular symmetry by changing to cylindrical coordinates: r, $\theta$, z.
The largest extent parallel to the xy-plane is at the intersection of the two parabolas, a circle with radius 1 so r ranges from 0 to 1, $\theta$ varies from 0 to [itex]2\pi[/tex] and z varies from 16r2 to 32- 16r2.
The volume is
$$\int_{r=0}^1 \int_{\theta=0}^{2\pi} \int_{z= 15r^2}^1 r dzd\thetadr$$.