Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis: y=x^{2/3}, x=1 and y=0.

Answer 1

what i have is \[\pi \int\limits_{0}^{1}1^2-(y^{3/2})^2 dy\]

Answer 2

What is the axis of rotation?

Answer 3

and rotating it around the y-axis i would use washer method

Answer 4

i have tried washer and shell method, neither gave me the right answer, and i dont know how to plug this into wolframalpha to see what the right answer is

Answer 5

What is the right answer.

Answer 6

\[\pi \int\limits_{a}^{b} R^2-r^2 dr \] that is the washer method
\[2\pi \int\limits_{a}^{b} r \times h \] that is the shell method

Answer 7

i dont know what the right answer is... i dont know how to plug this into wolframalpha to find out what the right answer is

Answer 8

Why don't you integrate them?

Answer 9

I'm getting \(3\pi/4\)

Answer 10

pellet that is the right answer,@wio, how did you find that?

Answer 11

\[
\pi \int\limits_{0}^{1}1^2-(y^{3/2})^2 dy =\pi \int\limits_{0}^{1}1-y^3 dy =\pi\left( y-\frac{1}{4}y^4\bigg|_0^1\right) = \pi\left(1-\frac{1}{4}\right) =\frac{3\pi}{4}
\]

Answer 12

that is what i had but there was an error when i squared y^3/2 i wrote on my paper y^2 not y^3 and kept trying to integrate with y^2....

Answer 13

thank you very much @wio i know how to do these, fairly easy, but this one i kept getting wrong and kept overlooking my error...

Answer 14

\[
2\pi \int\limits_{a}^{b} r \times h = 2\pi \int\limits_{0}^{1} x \times x^{2/3}dx
=2\pi \int\limits_{0}^{1} x^{5/3}dx = 2\pi \left(\frac{3}{8}x^{8/3}\bigg|_0^1 \right)=\frac{3\pi}{4}
\]

Answer 15

Thank you @wio i will write this down in my notes so i have something to come back to on shell method

Answer 16

I found the answer on this link and they explained it a diff way and got a diff answer. http://andrewmath.com/m180/ApplicationIntegration/VolumesDiskWasher.pdf