Question

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

Answer 1

check ur question again plz

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

using double integral

Answer 7

hmm i think im getting it

Answer 8

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

Answer 9

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

Answer 10

i dont know for sure. i think points

Answer 11

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

Answer 12

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.

Answer 13

Actually, I should have said the washer method.

Answer 14

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