Chk Work/Answer - Volume of the solid obtained by rotating R around the line

Aug 2006
13
0
The region R is bounded by the graphs of \(\displaystyle x-2y=3\) and \(\displaystyle x=y^2\). Set up (but do not evaluate) the integral that gives the volume of the solid obtained by rotating R around the line \(\displaystyle x = -1\).

Area \(\displaystyle A = \pi(1+2y)^2 - \pi(1+y^2)2[/tex]

Volume \(\displaystyle V = \sum(\pi(1+2y)^2 - \pi(1+y^2)2*Dy\) = \(\displaystyle \int_{0}^{2}\pi(1+2y)^2 - \pi(1+y^2)^2\)\)
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, FLTR!

Your integral is slightly off . . .


The region R is bounded by the graphs of \(\displaystyle x-2y=3\) and \(\displaystyle x=y^2\).
Set up (but do not evaluate) the integral that gives the volume of the solid
obtained by rotating R around the line \(\displaystyle x = -1\).

Did you make a sketch?
Code:
          |
      :   |                 /
      :   |            ...o (9,3)
      :   |      ..*::::/
      :   |  .*:::::::/
      :   |*::::::::/
   ---+---*:-:-:-:/------------
    -1:   |*::::/
      :   |   o (1,-1)
      :   | /      *
      :   /               *
        / |

We will integrate with respect to \(\displaystyle y\) and use "washers".

The two functions are: .\(\displaystyle \begin{array}{cc}x\:=\:2y + 3 \\ x = y^2\end{array}\)

The "outer radius" is: .\(\displaystyle r_o\:=\:(2y + 3) -(-1) \:=\:2y + 4\)
The "inner radius" is: .\(\displaystyle r_i\:=\:y^2 - (-1) \:=\:y^2 + 1\)
. . And \(\displaystyle y\) ranges from \(\displaystyle -1\) to \(\displaystyle 3.\)

. . . . \(\displaystyle V \;=\;\pi\int^3_{-1} \bigg[(2y + 4)^2 - (y^2 + 1)^2\bigg]\,dy\)

 

galactus

MHF Hall of Honor
Jul 2006
3,002
1,124
Chaneysville, PA
It's fun to do these things both shells and washers. One is often times more difficult than the other, though.

Here's shells:

\(\displaystyle 2{\pi}\left[\int_{0}^{9}(x+1)(\sqrt{x}-\frac{x-3}{2})dx-\int_{0}^{1}(x+1)(-\sqrt{x}-\frac{x-3}{2})dx\right]\)

This method subtracts that tiny shaded area below \(\displaystyle -\sqrt{x}\) and above \(\displaystyle \frac{x-3}{2}\)

Which leaves the hash-marked shade left to rotate.
 
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