Should my volume be negative in this case

[Saladsamurai]

I have not run into a volume problem that returned a negative value as of yet...until now.

I have the following problems:
Find the volume when the region, in the FIRST quadrant,bounded by y=x^2; y=2-x; and x=0 is:
revolved around the x-axis and then revolved around the line y=2.(a) For the x-axis I got:
$$v=\pi\int_0^1[(2-x)^2-(x^2)^2]dx$$
$$=\pi\int_0^1[4-4x+x^2-x^4]dx$$
$$=\pi[4x-2x^2+\frac{x^3}{3}-\frac{x^5}{5}]^1_0=\frac{32\pi}{15}$$

(b) and for around y=2 I got
$$v=\pi\int_0^1[(x^2)^2-(2-x)^2]dx$$
$$=\pi[\frac{x^5}{5}-4x+2x^2-\frac{x^3}{3}]_0^1=-\frac{22\pi}{15}$$But I have my doubts about the latter.

Any help is appreciated,
Casey

 

[Saladsamurai]

How about this question: for part (b), am I using the right integral? To revolve around y=2, that is?

 

[Saladsamurai]

I give up. Usually Dick is yelling at me by now

 

[Saladsamurai]

Are my bounds reversed on part b? Since the line being rotated about is located "ABOVE" the region being rotated? I know it needs to be positive.

 

[Saladsamurai]

This one must be too tough or something.