Volume of a solid of revolution

[jason177]

Homework Statement

Find the volume of the solid that results when the region bounded by the curves y=16-x$$^{2}$$ and y=16-4x is revolved around x=8.

If you could show me how to do it with both the shell method and washer method it would be greatly appreciated.

Homework Equations

The Attempt at a Solution

When trying the shell method i tried to shift the graph to the left 8 units and then revolving it about the y-axis and got:
y=16-(x+8)$$^{2}$$ and y = 16-4(x+8) which gave me


2$$\pi$$*$$\int^{-4}_{-8}$$x(16-(x+8)$$^{2}$$ - (16-4(x+8))) dx

and that gave me a negative answer so I know that can't be right

 

[Mark44]

I would suggest working with the graphs as they are. Your way could work, but seems prone to errors that might not otherwise occur, and doesn't really simplify things that I can see.

If you use washers you will need to solve each of your equations for x in terms of y. The volume of a typical volume element is
pi(R^2 - r^2)dy

For R, the larger radius, measure from the vertical line to the x value on the slanted line. For r, the smaller radius, measure from the vertical line to the x value on the parabola. For example, R = 8 - <x value on slanted line>. And similar for r, the smaller radius. Since R > r, then R^2 > r^2, so you'll get a positive value for V.

Using shells, the volume of the typical volume element is
2*pi*R*(y val on parabola - y val on slanted line)*dx. R is the radius of the shell, measured from the line x = 8. If you measure it the right way you should get a positive value for V.