1. ## Shell Method (help)

Hello. First, I would like to say that this is my first post so thank you in advance for any help. I have tried numerous ways of setting this problem up and I have not been able to come up with the correct answer. Also, I'll try to figure out how to post with all the correct symbols, but for now I will just do the best I can with what my keyboard has on it.

The problem tells me to use the Shell Method to find the volume of the solid generated by revolving the region about the given axis.

The region bounded by y=5x^(1/2) , y=5 , and x=0 , about the line x=1.

Here is one of the ways I have tried setting this up.

Integral symbol = $After finding 5x^(1/2)=5 = x=1 I Integrate on the x-axis from 0-1$2pi[shell radius][shell height]dx

=2pi\$[1-x][5x^(1/2)]dx

I come up with 8pi/3 but the answer sheet says I am supposed to get 7pi/3.

I applogize again for the shabby way my post looks. I'm a little embarrassed after looking at some other peoples questions.

2. Hello, Quasar!

Welcome aboard!

Your set-up is almost correct . . .

Use the Shell Method to find the volume of the solid:
the region bounded by $y = 5x^{\frac{1}{2}},\:y=5,\:x=0$, about the line $x=1.$

Here is one of the ways I have tried setting this up.

After finding: $5x^{\frac{1}{2}}= 5\quad\Rightarrow\quad x=1$
. . I Integrate on the x-axis from 0 to 1.

$V \;=\; 2\pi\int^1_0[\text{shell radius}][\text{shell height}]\,dx$ . . . Right!

. . $= \;2\pi\int^1_0(1-x)\underbrace{(5x^{\frac{1}{2}})}_{\text{no}}\,dx$

Did you make a sketch?
Code:
      |
|         :
5+ - - - - *
|:::::*   :
|::*      :
|*        :
- * - - - - + - - - - - -
0         1

The height is: $5 - 5x^{\frac{1}{2}}$

3. ## Thank You

Thank you Soroban. Your shell height was right on. It seems silly that I overlooked that but sometimes I get stuck chasing my tail and can't see what is right in front of me. I have also figured out how use Microsoft Word to make equations, so anything I post in the future should look a little better.

Quasar