The region R is bounded by the graphs of and . Set up (but do not evaluate) the integral that gives the volume of the solid obtained by rotating R around the line .
Area [Math]A = \pi(1+2y)^2 - \pi(1+y^2)2[/tex]
Volume =
The region R is bounded by the graphs of and . Set up (but do not evaluate) the integral that gives the volume of the solid obtained by rotating R around the line .
Area [Math]A = \pi(1+2y)^2 - \pi(1+y^2)2[/tex]
Volume =
Hello, FLTR!
Your integral is slightly off . . .
The region R is bounded by the graphs of and .
Set up (but do not evaluate) the integral that gives the volume of the solid
obtained by rotating R around the line .
Did you make a sketch?Code:| : | / : | ...o (9,3) : | ..*::::/ : | .*:::::::/ : |*::::::::/ ---+---*:-:-:-:/------------ -1: |*::::/ : | o (1,-1) : | / * : / * / |
We will integrate with respect to and use "washers".
The two functions are: .
The "outer radius" is: . 2y + 3) -(-1) \:=\:2y + 4" alt="r_o\:=\2y + 3) -(-1) \:=\:2y + 4" />
The "inner radius" is: .
. . And ranges from to
. . . .
It's fun to do these things both shells and washers. One is often times more difficult than the other, though.
Here's shells:
This method subtracts that tiny shaded area below and above
Which leaves the hash-marked shade left to rotate.