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

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

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 $$A = \pi(1+2y)^2 - \pi(1+y^2)2$$

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

2. 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\:=\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 $3.$

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

3. 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.