Use a triple integral to find the volume of the solid bounded.

• April 22nd 2010, 03:16 PM
asdf122345
Use a triple integral to find the volume of the solid bounded.
z=9-x^3, y=-x^2+2, y=0, z=0, x is greater than and equal to 0.
• April 22nd 2010, 03:57 PM
AllanCuz
Quote:

Originally Posted by asdf122345
z=9-x^3, y=-x^2+2, y=0, z=0, x is greater than and equal to 0.

The definition of a triple integral is

$\int_a^b dx \int_{g_1(x)}^{g_2(x)} dy \int_{h_1(x,y)}^{h_2(x,y)} dz$

By following this we get,

$\int_0^{\sqrt{2}} dx \int_0^{-x^2 + 2} dy \int_0^{9-x^3} dz$
• April 22nd 2010, 04:06 PM
asdf122345
Where the radical x come from?
• April 22nd 2010, 04:11 PM
AllanCuz
Quote:

Originally Posted by asdf122345
Where the radical x come from?

$y=-x^2+2$

This gives our XY domain. To find our max value of x, set y=0. If you re-arrange such that

$x^2 = 2 - y$

Clearly, x is max when y=0.
• April 22nd 2010, 05:23 PM
asdf122345