1. ## cylindrical shells about the x-axis almost got it

I'm having a hard time with this, I'm getting (804pi)/5 which I'm sure is wildly inaccurate

which step is wrong?
x = (y+2)^3 + 1
new limits are 0 to 2
integrate:
2 pi (integral from 0 to 2){ ((y+2)^3 +1)(y) dy }
where y is the radius and (y+2)^3 +1 is the height

there's a very similar question in the textbook with nearly the same numbers that comes out to like (81pi)/5 so I know I'm on the right track

2. ## Re: cylindrical shells about the x-axis almost got it

Originally Posted by RamzaBeoulve

I'm having a hard time with this, I'm getting (804pi)/5 which I'm sure is wildly inaccurate

which step is wrong?
x = (y+2)^3 + 1
new limits are 0 to 2
integrate:
2 pi (integral from 0 to 2){ ((y+2)^3 +1)(y) dy }
where y is the radius and (y+2)^3 +1 is the height

there's a very similar question in the textbook with nearly the same numbers that comes out to like (81pi)/5 so I know I'm on the right track

3. ## Re: cylindrical shells about the x-axis almost got it

Originally Posted by alexmahone
well I tried the example in the book (I know that answer is right because it is given) but couldn't get that answer either
also this homework problem gets handed in online and the site instantly tells you whether or not it's right

4. ## Re: cylindrical shells about the x-axis almost got it

Originally Posted by RamzaBeoulve
well I tried the example in the book (I know that answer is right because it is given) but couldn't get that answer either
also this homework problem gets handed in online and the site instantly tells you whether or not it's right
Your solution seems right to me.

5. ## Re: cylindrical shells about the x-axis almost got it

Originally Posted by alexmahone
Your solution seems right to me.
would you mind doing another similar problem and reporting the answer you get?
if we both get the same answer, then I suppose the website is wrong and we're right.

obtain the volume using cylindrical shells of the following function when rotated about the x-axis
y=(x^(1/3)) -2
8<=x<=27

6. ## Re: cylindrical shells about the x-axis almost got it

$V = 2\pi \int_0^2 y[64-(y+2)^3] \, dy = \frac{496\pi}{5}$

confirmed the solution using disks ...

$V = \pi \int_9^{65} (\sqrt[3]{x-1}-2)^2 \, dx = \frac{496\pi}{5}$

7. ## Re: cylindrical shells about the x-axis almost got it

Originally Posted by skeeter
$V = 2\pi \int_0^2 y[64-(y+2)^3] \, dy$
How did you get the term inside the square brackets? I can see that it is equal to $65-x$ but why is it not simply $x$?

8. ## Re: cylindrical shells about the x-axis almost got it

we have almost the same formula except you have a (64-height) in yours
where does that come from?

9. ## Re: cylindrical shells about the x-axis almost got it

Originally Posted by alexmahone
How did you get the term inside the square brackets? I can see that it is equal to $65-x$ but why is it not simply $x$?
because integration is w/r to y

$y = \sqrt[3]{x-1} - 2$

$y+2 = \sqrt[3]{x-1}$

$(y+2)^3 = x-1$

$x = (y+2)^3 + 1$

horizontal "height" of the shell = right x-value - left x-value ...

$65 - x = 65 - [(y+2)^3 + 1] = 64 - (y+2)^3$

10. ## Re: cylindrical shells about the x-axis almost got it

obtain the volume using cylindrical shells of the following function when rotated about the x-axis
y=(x^(1/3)) -2
8<=x<=27
shells ...

$V = 2\pi \int_0^1 y[27 - (y+2)^3] \, dy$

confirm the integral by using disks about the x-axis ...

$V = \pi \int_8^{27} (\sqrt[3]{x} - 2)^2 \, dx$