Set up a definite integral that represents the volume obtained by rotating the region between the curve $y^{2}=x$ and $y^{2}=2(x-1)$ about the lines below:
(a) Rotate about the y-axis
(b) Rotate about the line $y=3$
The question itself is not hard but I'm having some trouble deciding as to how I should set up the integral. Any guidance?
On
$\textbf{a)}$ Using the disc method and symmetry, we have
$\hspace{.2 in}\displaystyle V=2\int_0^{\sqrt{2}}\pi((R(y))^2-(r(y))^2)dy=\color{blue}{2\int_0^{\sqrt{2}}\pi\left(\left(\frac{y^2}{2}+1\right)^2-(y^2)^2\right)dy}$
$\;\;\;$Alternate answer: (shell method)
$\hspace{.2 in}\displaystyle V=\color{green}{\int_0^{2}2\pi x\big(2\sqrt{x}\big)dx-\int_1^{2}2\pi x\big(2\sqrt{2(x-1)}\big)dx}$
$\textbf{b)}$ Using the shell method,
$\hspace{.2 in}\displaystyle V=\int_{-\sqrt{2}}^{\sqrt{2}}2\pi r(y) h(y)dy=\color{blue}{\int_{-\sqrt{2}}^{\sqrt{2}}2\pi (3-y)\left(\left(\frac{y^2}{2}+1\right)-y^2\right)dy}$
$\;\;\;$Alternate answer: (disc method)
$\hspace{.2 in}\displaystyle V=\color{green}{\int_0^{2}\pi\left((3+\sqrt{x})^2-(3-\sqrt{x})^2\right)dx-\int_1^{2}\pi\left(\left(3+\sqrt{2(x-1)}\right)^2-\left(3-\sqrt{2(x-1)}\right)^2\right)dx}$
When rotated around the y axis the integral is $V=\int \pi x^2 dy$.
in your case it would be $$\int \pi (x_2^2-x_1^2)dy$$ where $x_2=f(y),x_1=g(y)$ .
first of all you take a strip of width $dy$ around the $y$ axis and find the volume of that infinitesimal space generated.
Intuitively it becomes clear that the volume will be the area of the disk multiplied by the element $dy$ to get volume of revolution. $$\int^{\sqrt2}_{-\sqrt2} \pi(y^4-(y^2/2+1)^2)dy$$
b) rotating about the $y=3$ $$V=\pi\int^{2}_{0} ((3-\sqrt x)^2-(3-\sqrt{2x-1})^2)dx$$