How Do You Calculate the Volume of a Solid Bounded by z=8xy in the First Octant?

[dannyzambrano]

Find the volume of the solid in the first octant of xyz space, bounded below by the coordinate axes and the unity circle and bounded above by z = 8xy.

A) 1/2 B) 1 C) 2 D) 4 E) 8

I know definitely volume will be the double integral of 8xy dy dx.

I think my limits of integration for the inner integral should be -sqrt(1-x^2) to sqrt(1-x^2). Since we are looking at the unit circle ( x^2 + y^2 = 1)

The outer integral limits should just be from -1 to 1?

Is this correct?

When I do the inner integral I get 4x(y^2) evaluated between -sqrt(1-x^2) to sqrt(1-x^2) but this looks like it just gives me 0 when I do the inner integral...

4x(sqrt(1-x^2))^2 - 4x(-sqrt(1-x^2))^2

Can someone help me and tell me if I am doing something wrong. I think i am

 

[radou]

Your limits of integration are not correct.

Edit: both "inner" and "outer". Recall that you're in an octant.

 

[dannyzambrano]

oh right we are in the first octant.. so we are really only looking at a quarter of the unit circle i guess..

So my limits on the outer integral should just be from 0 to 1 right? and the limits on the inner integral should be from 0 to (sqrt(1-x^2))/2? since its just a quarter of the circle? Thats my reason for dividing by 2

 

[radou]

Yes, correct.

 

[invisible_man]

change to the sphendrical coordinate