So even after going to my profs office I could not find out what I did wrong in this simple integration rotation (washer) problem.
I'm given two curves x = y^2 and y = x^2 and I am instructed to rotate this along y=1. To start I convert x = y^2 into y = √x and set them equal to each other. I find that they meet at (1,1) and (0,0). Next I write out my area function considering the fact that if I create a washer to handle the problem my big R will equal (1 - x^2) and my little r will be
$$ r= (1 - √x) $$
$$ A(x) = π[(1 - x^2)^2 - (1 - √x)^2] = π[1 - x^4 - 2x^2 - 1 -x +2x^(1/2)] $$
$$= \int_0^1 \pi[- 1/5x^5 - 2/3x^3 + 1/2x^2 +4/3x^(3/2)] (1,0) $$
$$ = \pi[-1/5 - 2/3 + 1/2 + 4/3] = \pi[-26/30 + 55/30] = 29\pi/30 $$
However this is not the correct answer. I've stared at this problem for some time and I know someone will be able to quickly point me in the right direction! Sorry if my formatting is awful, I sincerely appreciate it. I also provided the correct graphs that I used to solve this problem.
Very simple error may be, help from a second set of eyes would be greatly appreciated!
With the method of washers, the inner radius of the solid is given by $$g(x) = 1-\sqrt{x}, \quad 0 \le x \le 1.$$ The outer radius is $$f(x) = 1 - x^2, \quad 0 \le x \le 1.$$ Then, at a given $x$-value, a washer of differential thickness $dx$ has differential volume $$dV = \pi(f(x)^2 - g(x)^2) \, dx = \pi ((1-x^2)^2 - (1 - \sqrt{x})^2) \, dx.$$ It is perhaps simpler to observe that $$f(x)^2 - g(x)^2 = (f(x) + g(x))(f(x) - g(x)),$$ so that $$f(x) - g(x) = (1-x^2) - (1-\sqrt{x}) = x^{1/2} - x^2, \\ f(x) + g(x) = 2 - x^{1/2} - x^2,$$ so the integral to be evaluated is $$V = \int_{x=0}^1 dV = \pi \int_{x=0}^1 (x^{1/2} - x^2)(2 - x^{1/2} - x^2) \, dx.$$ Now expanding the product, we get $$V = \pi \int_{x=0}^1 2x^{1/2} - x - 2x^2 + x^4 \, dx.$$ Integrating term by term gives $$V = \pi \left[ \frac{4}{3}x^{3/2} - \frac{x^2}{2} - \frac{2}{3}x^3 + \frac{x^5}{5} \right]_{x=0}^1 = \pi \left(\frac{4}{3} - \frac{1}{2} - \frac{2}{3} + \frac{1}{5} \right) = \frac{11\pi}{30}.$$