whats the volume of y=x^(2/3) , x=1, y=0; about the y-axis?

Question

anonymous · Answer

check ur question again plz

anonymous · Answer

whats the volume of y=x^(2/3) , x=1, y=0; rotated about the y-axis?

anonymous · Answer

i got 3pi/4, but i'm not sure

anonymous · Answer

hmm i got pi/4 the integral part i had i think is wrong i got from 0 to 1

anonymous · Answer

actually in this case, xrun from 0 to 1, y run from 0 to x^(2/3), so i think we can calculate the area, then multiply by 2pi to find volume

anonymous · Answer

using double integral

anonymous · Answer

hmm i think im getting it

anonymous · Answer

hey dude, why u mean by x=1, y=0???
Are these a point or a vector???

anonymous · Answer

hey dude, what u mean by x=1, y=0??? Are these a point or a vector???

anonymous · Answer

i dont know for sure. i think points

anonymous · Answer

actually those numbers are used to find the $$?\le x \le ?$$

anonymous · Answer

I rewrote as functions of x and used the disk method.
$$\pi \int\limits_{0}^{1} (1)^2-(y^(2/3))^2 dy$$
$$=\pi \int\limits_{0}^{1} 1-y^3 dy$$
$$= 2\pi/3$$
But, I wouldn't bet my life on it.

anonymous · Answer

Actually, I should have said the washer method.

anonymous · Answer

I made a mistake evaluating the integral.  The answer should be $$3\pi/4$$