Volume of Rotation: Find y3=x2 Volume in 64π units^3

[danago]

Find the volume of the solid generated by rotating the region trapped between the curve y³=x², the y-axis, the line y=4 and the line x=0 around the y-axis.

I started by writing x as a function of y, explicitly:

$$x=y^{1.5}$$

Heres the graph i obtained, with the shaded area being the area to be rotated about the y axis.

http://img241.imageshack.us/img241/1429/135q3hot9.png

$$V = \pi \int\limits_0^4 {(y^{1.5} )^2 dy = } \pi \int\limits_0^4 {y^3 dy = } 64\pi {\rm{ units}}^3$$

The answer in the book says it should be 631.65 units³. It looks to me as if they multiplied by $$\pi^2$$ instead of just $$\pi$$. Am i missing something, or am i on the right track?

Thanks in advance,
Dan.

 

[Dick]

It looks like 64*pi units to me as well. Are we both missing something?

 

[rocomath]

i got the same answer and i haven't solved any volume problems in a long time

 

[Dick]

That concludes it. The score is 3 against 1. The book answer is wrong. Not all that unusual.

 

[danago]

Alright that's good to hear Thanks for confirming it guys

 

[P.O.L.A.R]

Is it a solutions manual or the back of the book?

They did square pi but I don't see how they did. I wonder if they thought that pi should be squared because the radius (R(x)) is squared which is completely wrong. Quite strange really.