Question

Find the volume of the solid obtained by rotating the region bounded by \(y = \sqrt[3]{x}\), y = 0 and x = 1 about the y-axis.

I just want to find out how to set this up in the form \(\int_a^b 2\pi r h dx\) so i can find a pattern in finding them =_=

Note: no inkyvoyds allowed :p

Answer 1

@inkyvoyd Any ideas?

Answer 2

you are setting it up correctly, and because its rotating around the y-axis, your y value is the height and your x values (the bounds) are the radius in either direction
2pi is a constant so remove it from integration before integrating, then solve for dx from b to a (your bounds)

Answer 3

im thinking \[2\pi \int_0^1 y^3 dy\] that right?

Answer 4

my r = 1

Answer 5

hmmm, 'a' might be -1 and 'b' might be 1,
cause the bounds are from -1 to 1 since r=1 in both directions from the origin.
i'll check real quick

Answer 6

ohhhh

Answer 7

oh wait... 0 to 1 is with respect to the y-axis...

Answer 8

Volume should equal \[pi \int\limits_{a}^{b}[f(x)]^2dx\]
when rotating around the x axis.
since you are rotating around the y axis im having a hard time picturing for dy

Answer 9

i think that's circular disk?

Answer 10

About y-axis => shell method!!~~

Answer 11

yeah...i was asking which i'll use as \[\int_a^b 2\pi rhdx\] i posted my substitutions up tehre...and is it automatic shell when about y-axis??

Answer 12

r=x, h = f(x)

Answer 13

Try rearranging y = x^1/3 to y^3 = x and take the integral with respect to y from 0 to 1.

Answer 14

uhmmm yeah..i used

\[\int_0^1 2\pi (1) y^3 dy\]

r = 1

Answer 15

0<=x<=1
0<=y<=cube root of x so you are to use
\[\int\limits_{0}^{1} pi (\sqrt[3]{x})^2dx\]
not the one you just typed :D