# Triple Integral

• Mar 26th 2010, 11:52 AM
ur5pointos2slo
Triple Integral
Find the volume of the region.

The region between the cylinder z=y^2 and the xy-plane that is bounded by the planes x=0,x=1,y=-1,y=1

When I graph this it looks like a halfpipe(skateboarding) extending in the positive x direction.

my attempt:

int 0..1 int x^2+1.. x^2-1 int 0.. y^2 dz dy dx
• Mar 26th 2010, 01:02 PM
shawsend
That's just a double integral. You're integrating from the x-y plane up to the surface that is $\displaystyle z=f(x,y)=y^2$. That's the z-direction. And we're integrating that over the foot-print in the x-y plane that's just the two unit squares underneath the surface which because it's symmetric, just multiply by two, the volume underneath the unit square as x goes from 0 to 1 and y goes from 0 to one or:

$\displaystyle V=2\int_0^1\int_0^1 y^2 dydx$

or if you wanna' be a purist, a triple integral:

$\displaystyle V=2\int_0^1\int_0^1\int_0^{y^2} dzdydx$

but same dif.
• Mar 26th 2010, 01:12 PM
ur5pointos2slo
Quote:

Originally Posted by shawsend
That's just a double integral. You're integrating from the x-y plane up to the surface that is $\displaystyle z=f(x,y)=y^2$. That's the z-direction. And we're integrating that over the foot-print in the x-y plane that's just the two unit squares underneath the surface which because it's symmetric, just multiply by two, the volume underneath the unit square as x goes from 0 to 1 and y goes from 0 to one or:

$\displaystyle V=2\int_0^1\int_0^1 y^2 dydx$

or if you wanna' be a purist, a triple integral:

$\displaystyle V=2\int_0^1\int_0^1\int_0^{y^2} dzdydx$

but same dif.

I will take the purist direction. The only reason is that it is under the triple integral section. Thanks for the help.