Thread: Volume of a solid generated by rotating a region about the y-axis

1. Volume of a solid generated by rotating a region about the y-axis

"Sketch the region bounded by the curves $\displaystyle y = x^ {\frac{3}{2}}$, y = 8, x = 0, and find the volume of the solid generated by revolving the region about the y-axis."

I've sketched the region, and can visualise what they're looking for with the rotation thing, but I don't know how one actually calculates it...

I'd suspect integration is needed, but I'm not sure exactly what I'm supposed to be integrating...

2. Originally Posted by PTL
"Sketch the region bounded by the curves $\displaystyle y = x^ {3}{2}$, y = 8, x = 0, and find the volume of the solid generated by revolving the region about the y-axis."

I've sketched the region, and can visualise what they're looking for with the rotation thing, but I don't know how one actually calculates it...

I'd suspect integration is needed, but I'm not sure exactly what I'm supposed to be integrating...
did you mean for your function to be $\displaystyle y = x^{\frac{3}{2}}$ ?

3. Yep, sorry.. Will edit it.

4. Originally Posted by PTL
"Sketch the region bounded by the curves $\displaystyle y = x^ {\frac{3}{2}}$, y = 8, x = 0, and find the volume of the solid generated by revolving the region about the y-axis."

I've sketched the region, and can visualise what they're looking for with the rotation thing, but I don't know how one actually calculates it...

I'd suspect integration is needed, but I'm not sure exactly what I'm supposed to be integrating...
two methods.

cylindrical shells ...

$\displaystyle V = 2\pi \int_0^4 x(8 - x^{\frac{3}{2}}) \, dx$

disks ...

$\displaystyle V = \pi \int_0^8 y^{\frac{4}{3}} \, dy$

5. Ok, I think the disks one seems simpler.. I'll go with it..

I see you're using pi r^2 and stacking the disks on top of one another, from 0 to 8 along the y-axis, that's fine - but where does the y^{4/3} come from? I presume it's something to do with the r^2... but I don't see where it comes from...

6. Originally Posted by PTL
Ok, I think the disks one seems simpler.. I'll go with it..

I see you're using pi r^2 and stacking the disks on top of one another, from 0 to 8 along the y-axis, that's fine - but where does the y^{4/3} come from? I presume it's something to do with the r^2... but I don't see where it comes from...
the radius of each disk is $\displaystyle x$

$\displaystyle y = x^{\frac{3}{2}}$

$\displaystyle x = y^{\frac{2}{3}}$

$\displaystyle r^2 = x^2 = \left(y^{\frac{2}{3}}\right)^2 = y^{\frac{4}{3}}$

7. Originally Posted by skeeter
the radius of each disk is $\displaystyle x$

$\displaystyle y = x^{\frac{3}{2}}$

$\displaystyle x = y^{\frac{2}{3}}$

$\displaystyle r^2 = x^2 = \left(y^{\frac{2}{3}}\right)^2 = y^{\frac{4}{3}}$
Ah, of course - I was thinking {2/3}^2 = 4/9. Gotcha.

So integrating you get 3/7 * y ^ {7/3} between 0 and 8, gives 54.8 pi

Thanks