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

#### FLTR

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
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
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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