calculusintegrationvolume
The region between $y = sin(x^2)$ and the x-axis for $0 \le x \le \sqrt{\pi}$ is revolved around the y-axis. Find the volume of the resulting solid.
I can get all the way to the integral: $ \pi\int_0^\sqrt{\pi} sin(x^2)dx$ but do not know how to proceed. Is this correct so far?
I assumed the following colored are is supposed to revolve about $x$ axis.:
$x=1$ gives $y=\sqrt{2}$, so we have $$V=2\pi\int_{y=0}^{\sqrt{2}}y(x_2-x_1)dy$$ where $y^2-1=x_2,~~\frac{y^2}{2}=x_1$.
We need to rotate about the line $x=0$ (y-axis), and we want to express $x$ as a function of y: $x=1-y^2$ (as given). That's a horizontal parabola, intersecting the y-axis at $y =-1$ and $y = 1$. The aim here is to integrate with respect to $y$, where $x = 1-y^2$ is the radius, which varies as $y$ varies.
$$\begin{align} \pi \int_{-1}^1 (1-y^2)^2\,dy & = \pi \int_{-1}^1 (1- 2y^2 + y^4)\,dy\\ & = \pi\left(y - \frac {2y^3} 3 + \frac{y^5}5\right)\Bigg|_{-1}^1 \end{align}$$ I'll let you take it from here: just be careful with the signs.
Best Answer
By the shell method, $$dV=2\pi x y dx\\ \implies dV=2\pi x\sin(x^2)dx\\ \implies V=2\pi\int^\sqrt{\pi}_{0}x\sin(x^2)dx\\ \implies V=\pi\int^\sqrt{\pi}_{0}\sin(x^2)d(x^2)\\ \implies V=-\pi\cos(x^2)|^\sqrt{\pi}_0\\ \implies V=-\pi\cos(\pi)+\pi\cos(0)=\pi+\pi=2\pi$$