1. ## Check answer for volume by revolution

Can you please tell me if i worked the following out correctly?

$y=x^2$
$y=x+2$

1. Rotate this around the x-axis
outside radius = $x+2$
inside radius = $x^2$

$\pi\int_1^2 (x+2-x^2) = \frac{9\pi}{2}$
(thats suppose to be -1 to 2 but i cant get latex to work)
2. Rotate around $y = 4$
outside radius = $4-x^2$
inside radius = $4-(x+2)$

$\pi\int_1^2 (4-x^2)-(4-(x+2)= \frac{9\pi}{2}$
(thats suppose to be -1 to 2 but i cant get latex to work)
3. Rotate around x = -2
outside radius = $\sqrt{y}+2$
inside radius = $2-y+2$

$\pi\int_0^4 (\sqrt{y}+2-(2-y+2)= \frac{16\pi}{3}$

2. Originally Posted by silencecloak
Can you please tell me if i worked the following out correctly?

$y=x^2$
$y=x+2$

1. Rotate this around the x-axis
outside radius = $x+2$
inside radius = $x^2$

$\pi\int_1^2 (x+2-x^2) = \frac{9\pi}{2}$
(thats suppose to be -1 to 2 but i cant get latex to work)

Put the -1 in braces: \int_{-1}^2.
The area of a circle is $\pi r^2$ you should have $(x+2)^2- (x^2)^2$.

2. Rotate around $y = 4$
outside radius = $4-x^2$
inside radius = $4-(x+2)$

$\pi\int_1^2 (4-x^2)-(4-(x+2)= \frac{9\pi}{2}$
(thats suppose to be -1 to 2 but i cant get latex to work)

Again, the radii should be squared.

3. Rotate around x = -2
outside radius = $\sqrt{y}+2$
inside radius = $2-y+2$

$\pi\int_0^4 (\sqrt{y}+2-(2-y+2)= \frac{16\pi}{3}$

same comment for the last one!

1. $72\pi/5$

2. $108\pi/5$

3. $24\pi$

Are these correct?

4. Im not 100% confident in my set up can anyone please confirm?

Thank you!

5. I solved for the first two answers and also got
and .
However, I have not solved for the 3rd answer, as I need to verify my methods involving negative axes of revolution.
I am sure of my first two answers, though.

6. Ya my main concern is number 3 as well

7. Anyone else have an opinion?

8. 3) I couldn't understand how did you applied the disk method so I used shell method because i think it's easier in this case:

$V=2\pi\int_{-1}^2(x+2)(x+2-x^2)dx=2\pi\int_1^4t(t-(t-2)^2)dt=\frac{45\pi}{2}.$