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

**FLTR** · August 31st 2006, 05:52 PM

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

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

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

**Soroban** · August 31st 2006, 06:39 PM

Hello, FLTR!

Your integral is slightly off . . .

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

Did you make a sketch?

Code:

          |
      :   |                 /
      :   |            ...o (9,3)
      :   |      ..*::::/
      :   |  .*:::::::/
      :   |*::::::::/
   ---+---*:-:-:-:/------------
    -1:   |*::::/
      :   |   o (1,-1)
      :   | /      *
      :   /               *
        / |

We will integrate with respect to $y$ and use "washers".

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

The "outer radius" is: . $r_o\:=\<img src=$ 2y + 3) -(-1) \:=\:2y + 4" alt="r_o\:=\

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

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

**galactus** · September 1st 2006, 04:37 AM

It's fun to do these things both shells and washers. One is often times more difficult than the other, though.

Here's shells:

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

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

Which leaves the hash-marked shade left to rotate.

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