Question

y=x^(1/3) surface area of revolution from y=1 to 2.

edit: revolved around y axis

Answer 1

\[2\pi \int\limits_{a}^{b}\sqrt{1+[f'x]^2}xdx\]

Answer 2

f'x=.33x^-.666
[f'x]^2=.1111/x^1.3333 which is equal to 1/(9x^1.3333)

Answer 3

revolved about the x-axis or about the y-axis?

Answer 4

y axis

Answer 5

the equation i have should be right

Answer 6

\[\Large 2\pi\int_{y_1}^{y_2}x\sqrt{1+(x')^2}\;\mathrm dy\]

Answer 7

\[\Large y=x^{1/3}\Rightarrow y^3=x\]

Answer 8

it should be not difficult to integrate

Answer 9

ok. for the x in the equation, do i just pu x or y^3

Answer 10

replace x by y^3

Answer 11

o yay. i got it, thanks for the help.

Answer 12

yw