Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
$y=x^2, x = y^2;$ rotate about $y=1$.
So the first thing that jumps out to me is that we're rotating this around $y=1$. So it seems like using vertical washers might do the trick. So I'm going to represent all the equations in terms of x. So $y = x^2$ and $y = \sqrt{x}$
radius outer: $1 - x^2$ radius innter: $1 - \sqrt{x}$
$$ \pi \int_0^1 (1 - x^2)^2\,dx - \pi \int_0^1 (1 - \sqrt{x})^2 \,dx$$
$$ = \pi \int_0^1 (1 - 2x^2 + x^4) \,dx - \pi \int_0^1 (1 - 2\sqrt{x} + x)\,dx$$
$$ = \pi \int_0^1 (-2x^2 + 2\sqrt{x} + x^4 - x)\,dx$$
Is this setup right so far? Is the answer $\frac{\pi * 11}{30}$
You've done very well. Your work is correct, and indeed, the answer is $\frac{11\pi}{30}$.
Now it's simply a matter of using the power rule for integration, evaluating at $x=1$, and $x=0$, finding a common denominator. I edited your post to include the missing $dx$ in each integral.
$$\pi \int_0^1 (-2x^2 + 2\sqrt{x} + x^4 - x)\,dx = \pi\int_0^1(-2x^2 + 2x^{1/2} + x^4 - x^1)\,dx $$ $$= \left(-\frac 23 x^3 + \frac 43x^{3/2} + \frac 15 x^5- \frac 12x^2\right) \Big|_0^1= \frac {11\pi}{30}$$