# Thread: Calculus: volumes of rotation

1. ## Calculus: volumes of rotation

i need some help w/ this problem

the region bounded by y=x^3, y=0, x=3. Is rotated about the line x=-1. What is the volume of the resulting solid?

help would greatly be appreciated

2. Originally Posted by bavery25
i need some help w/ this problem

the region bounded by y=x^3, y=0, x=3. Is rotated about the line x=-1. What is the volume of the resulting solid?

help would greatly be appreciated
Hint: use the shell method.

can you tell what the radius and height are?

3. Hello, bavery25!

Did you make a sketch?

The region bounded by $\begin{Bmatrix}y\:=\:x^3 \\y\:=\:0 \\ x\:=\:3\end{Bmatrix}$ is rotated about the line $x=-1$

What is the volume of the resulting solid?
Code:
              |
|          *
|
:     |         *
:     |        *|
:     |      *::|
:     |   *:::::|
- - : - - * - - - - + - -
-1     |         3
|

Formula: . $V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx$

So we have: . $V \;=\;2\pi\int^3_0(x+1)(x^3)\,dx$

4. (1+x) is radius i think and height is (x^3)

5. oops need to correct it its x=1 not x=3 sry

6. Originally Posted by bavery25
oops need to correct it its x=1 not x=3 sry
no matter, just change the 3 in Soroban's post to a 1, the upper limit of the integral

7. yea i had that part but my professor has the answer as 7pi/6 and i cant get that so idk what im doing wrong

8. Originally Posted by bavery25
yea i had that part but my professor has the answer as 7pi/6 and i cant get that so idk what im doing wrong
i don't get $\frac {7 \pi}6$ either. make sure there's not some other problem with the question

9. nope thats the question with the problem so i guess he is wrong

10. what did u get as the answer