definite integralsvolume
The region between curves $y= {\sqrt x}$ , $0≤x≤4,y=1,x=4$ is revolved about $y=1$. Find the volume of a generated solid.
I believe I need to find volume outlined by green, blue and dotted, black line (like on the graph).
$V = \pi $$\int_1^4 ({\sqrt x})^2 \,dx$. Is it correct approach or should I calculate $\pi $$\int_0^4 ({\sqrt x})^2 \,dx$ instead or maybe there is different approach.
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 $y^2-1=x_2,~~\frac{y^2}{2}=x_1$.
By the shell method, $$dV=2\pi x y dx\\ \implies dV=2\pi x\sin(x^2)dx\\ \implies V=2\pi\int^\sqrt{\pi}_{0}x\sin(x^2)dx\\ \implies V=\pi\int^\sqrt{\pi}_{0}\sin(x^2)d(x^2)\\ \implies V=-\pi\cos(x^2)|^\sqrt{\pi}_0\\ \implies V=-\pi\cos(\pi)+\pi\cos(0)=\pi+\pi=2\pi$$
Best Answer
Hint: You have to calculate $$\pi\int_1^4(\sqrt{x}-1)^2dx$$ which is the volume of the curve $\sqrt{x}-1$ revolved around $y=0$ instead.