Question

find the surface are of the surface generated by revolving the curve
x=3/8(y4/3−2y2/3)
from y=1 to y=8 revolving about the x-axis.
so far this is what i have
Is any of this correct?
∫183/8(y4/3−2y2/3)
3/8[3/7y7/3−2(3/5)5/3]from8→1
3/8[(3/7)(8)7/3−2(3/5)(8)5/3]−3/8[(3/7)(1)7/3−2(2/5)(1)5/3]
(3/8)[(387/7)−(192/5)]−(3/8)[(3/7)−(6/5)]
(1773/280)−(291/280)
=740/140

Answer 1

its correct so farrr

Answer 2

For surface area it is:
\[2 \pi \int\limits_a^b r(y)\sqrt{1+(\frac{dx}{dy})^2}dy\]

Answer 3

In this case r(y)=y