1) Rotate the graph 4x^5 about the line x = 1 and calculate the volume of the generated figure.

I assumed that the radius $\displaystyle r = 1 - 4x^5$ and then integrated from 0 to 1 and was going to double it

$\displaystyle A = \pi (1-4x^5)^2$

$\displaystyle A = \pi (1-8x^5+16x^{10}$

$\displaystyle \int^1_0 A(x)dx = x-\frac{4}{3}x^6 + \frac{16x^{11}}{11}|^1_0$

Which gives you $\displaystyle 1-\frac{4}{3}+\frac{1}{11}$

2) $\displaystyle y = 5x, y = 5\sqrt{x}$ Rotated about y = 5. Find the volume

Washer method:

$\displaystyle A(x) = \pi (5-5x)^2-\pi (5-\sqrt{x})^2$

$\displaystyle \pi \int^1_0 (25-50x+25x^2)-(25-50\sqrt{x}-25x)$

$\displaystyle \pi \int^1_0 -75x+25x^2+50\sqrt{x}$

But that evaluates to 0, which doesn't make sense (and is also wrong).

Where did I go wrong on these?