Method of cylindrical shells
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.
$y=32-x^2, \ y=x^2$ about the line $x=4$
My confusion is that what will be the radius 'x' of cylinder shells which we have to put in the integral.
2 Answers
$\begingroup$Let $-4\le x\le 4$. Draw a thin vertical strip of width "$dx$" at $x$. For the picture, let $x$ for example be $1.5$.
Rotate this thin strip about the line $x=4$. We get a cylindrical shell. This shell has height $(32-x^2)-x^2$. Its distance from the line $x=4$ is $4-x$. That is the radius of the cylindrical shell. It follows that the volume is equal to $$\int_{-4}^4 2\pi(4-x)(32-2x^2)\,dx.$$
$\endgroup$ 2 $\begingroup$Cylindrical shell: consider the volume element $$ dV = 2 \pi h\,dr = 2\pi\,2y\,dx = 4\pi\,x^2dx $$ Then integrate $$ V = \int_{x=-4}^4 dV = 4\pi \int_{-4}^4 x^2dx $$
Cylindrical disk: Cut your solid in half so that you only have to consider the bottom part. Then integrate the volume element $$ dV = \pi x^2 dy = \pi (2\sqrt{y})^2 dy $$ $$V = \int_{y=0}^{16} dV = 4\pi \int_0^{16} y\,dy $$
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