The function is:
$y=\sin(x^2)$ and the boundaries are $y=0$, $x=0$, and $x=\sqrt{\pi}$
The function is rotated around the y-axis, and since the functions are difficult to express in terms of y and integrate with respect to dy, I use the formula of
$$V=2\pi \int^\sqrt{\pi}_0 \sin (x^2)dx$$
My two questions here are:
1) What method can I use to solve the integral of $\sin (x^2)$?
2) How do I limit the boundary in some way to not include the area below the x-axis?
As Doug M points out in the comments, to find the volume by the shell method, as explained here, we must use $$V= \int_{0}^{\sqrt{\pi}} (2\pi x)\sin(x^2)dx$$ where $2\pi x$ is the radius of the shell and $\sin(x^2)$ is the height of the shell. Unlike your previous integral, this integral has a closed form, and has a straightforward solution by substitution.
So as for your first question, the integral of $\sin^2(x)$ is not expressible in closed form. But if you read up on the Fresnel Integral, you could write it as $\sqrt{\frac{\pi}{2}}S(\sqrt{2})$, where $S(x)$ is the Fresnel S Integral.
And as for your second question, $\sin(x) \geq 0$ for $0 \leq x\ \leq \pi$, as Doug pointed out in the comments. Thus we don't need to worry about any area below the $x$ axis.