Question

find the area of the surface generated by revolving the curve about the indicated axis. y=(x^3)/14 x=0 x=4. ive gotten it down to S=2pi (x^3)/14 squareroot 1+(9x^4)/196 but now i am stuck

Answer 1

I think i have to integrate again but i am not sure what to integrate

Answer 2

Around which axis?

Answer 3

x axis im sorry

Answer 4

any ideas'

Answer 5

Stay tuned

Answer 6

thank you

Answer 7

\[
2 \pi f(x) \sqrt{f'(x)^2+1}=\frac{1}{7}
\pi x^3 \sqrt{\frac{9 x^4}{196}+1}\\
\int \frac{1}{7} \pi x^3 \sqrt{\frac{9
x^4}{196}+1} \, dx=\frac{\pi \left(9
x^4+196\right)^{3/2}}{5292}
\]

Answer 8

and do i plug in 4 and 0 or do i need to find new limits of integration?

Answer 9

\[
\int_0^4 \frac{1}{7} \pi x^3
\sqrt{\frac{9 x^4}{196}+1} \,
dx=\frac{1132 \pi }{49}
\]

Answer 10

thank you so much!!!

Answer 11

No you plug in 4 and 0 and you substract.

Answer 12

yw

Answer 13

awesome!!!

Answer 14

@eliassaab I am not getting the same answer. what equation do i plug those numbers into