Find the volume of the solid of revolution obtained by revolving the graph of $y = \frac{6}{x^2}$ around the $x$-axis over the range $[1,4]$.
Draw a picture of the region. What should the radius be for this region?
From the radius, what is the area of each cross section?
The integral to find the volume is $\displaystyle \int_1^4 \pi \left(\frac{6}{x^2}\right)^2\ dx$
$\displaystyle \frac{189}{16}\pi$
Find the volume of the solid of revolution obtained by revolving the region enclosed between the graphs of $y = 3 - x$ and $y = x^2 - 3$ around the line $y = -5$.
Sketch out a picture. Draw in the segment that is being rotated and try to set up the integral.
This is a washer method problem. What is the inner radius? What is the outer radius?
What should the bounds be on the integral that you need to compute? (And how would you go about finding them?)
The integral is $\displaystyle \pi \int_{-3}^2 (3 - x +5)^2 - (x^2 - 3 + 5)^2\ dx$
$\displaystyle 250\pi$
Find the volume of the solid of revolution obtained by revolving the region in the first quadrant between the graph of $y = x^2 + 4$ and the line $y = 13$ around the $x$ axis.
What is the segment being rotated? What are the inner and outer radii?
What should the bounds on the integral be?
The integral should be $$ \pi \int_0^3 13^2 - (x^2 + 4)^2\ dx $$
$\displaystyle \frac{1692}{5} \pi $
Find the volume of the solid of revolution obtained by revolving the region in the first quadrant between the graph of $y = x^2 + 4$ and the line $y = 13$ around the $y$ axis.
What is the segment being rotated? What are the inner and outer radii?
What should the bounds on the integral be?
The integral should be $$ \pi \int_4^{13} (y-4) \ dy $$
$\displaystyle \frac{81}{2} \pi $
Find the volume of the solid of revolution obtained by revolving the region in the first quadrant between the graph of $y = x^2 + 4$ and the line $y = 13$ around the line $x = -1$.
What is the segment being rotated? What are the inner and outer radii?
What should the bounds on the integral be?
The integral should be $$ \pi \int_4^{13} (\sqrt{y-4} + 1)^2 - 1^2 \ dy $$
$\displaystyle \frac{153}{2} \pi $
Find the volume of the solid of revolution obtained by revolving the region between the graphs of $y = 2x$ and $y = x^2 - 4x$ around the line $y = -4$
Find the volume of the solid of revolution obtained by revolving the region between the graphs of $y = x^4 + 1$, $y = 1$ and $x=2$ around the line $x = -1$.