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marigirl
 one year ago
I don't quite understand this question
The Surface area of a solid is given by the formula
\[S=2\pi \int\limits_{a}^{b}f(x)\sqrt{1+(f'(x))^2}dx\]
The graph of \[f(x)=x^3\] is rotated between x=0 and x=a, where a>0 about the x axis. If the surface area is 4pi, find the value of a
marigirl
 one year ago
I don't quite understand this question The Surface area of a solid is given by the formula \[S=2\pi \int\limits_{a}^{b}f(x)\sqrt{1+(f'(x))^2}dx\] The graph of \[f(x)=x^3\] is rotated between x=0 and x=a, where a>0 about the x axis. If the surface area is 4pi, find the value of a

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marigirl
 one year ago
Best ResponseYou've already chosen the best response.0\[S=2\pi \int\limits_{0}^{a} x^3 \sqrt{1+9x^4} dx\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0ill post the ans on here, and if u could explain it please :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3So you're having trouble with the integration?

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0i dont quite get what to integrate,

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Oh you can do a u substitution u = 1+9x^4

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0so just integrate the whole thing normally..the f(x) and f'(x) parts were putting me off

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, that's just showing you what you plug into the integral
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