More volume of a washer cross section(integration)

[Neutrinogun]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

Homework Equations

$y = x^2, x = y^2$ about $y = -1$

The Attempt at a Solution

I tried using both cylindrical and washer methods - but for cylindrical I couldn't figure out what the shell radius would be, so I switched to washer.

$\pi\int_0^1(x^2+1)^2 - (√x+1)^2 dx$
$\pi\int_0^1 x^4 + 2x^2 - x - 2√x dx$
$\pi(x^5/5 + (2x^3)/3 - x^2/2 - (4x$^^3/2$))/3)$
$\pi(1/5 + 2/3 - 1/2 - 4/3)$
$-29\pi/30$

And I get a negative volume. Any ideas?
EDIT: Ah. I just noticed I subtracted in the wrong order - would that fix it?
Edit2: Also, would you mind helping me with using the cylindrical method for this? Because I wasn't able to narrow down the shell radius - there was always a little unknown portion.

 

[SammyS]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

Homework Equations

$y = x^2, x = y^2$ about $y = -1$

The Attempt at a Solution

I tried using both cylindrical and washer methods - but for cylindrical I couldn't figure out what the shell radius would be, so I switched to washer.

$\pi\int_0^1(x^2+1)^2 - (√x+1)^2 dx$
$\pi\int_0^1 x^4 + 2x^2 - x - 2√x dx$
$\pi(x^5/5 + (2x^3)/3 - x^2/2 - (4x$^^3/2$))/3)$
$\pi(1/5 + 2/3 - 1/2 - 4/3)$
$-29\pi/30$

And I get a negative volume. Any ideas?
EDIT: Ah. I just noticed I subtracted in the wrong order - would that fix it?
Edit2: Also, would you mind helping me with using the cylindrical method for this? Because I wasn't able to narrow down the shell radius - there was always a little unknown portion.

For the cylindrical method, the radius is y - (-1), which is y + 1 .