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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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

Mathematics
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\[S=2\pi \int\limits_{0}^{a} x^3 \sqrt{1+9x^4} dx\]
ill post the ans on here, and if u could explain it please :)
So you're having trouble with the integration?

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i dont quite get what to integrate,
Oh you can do a u substitution u = 1+9x^4
so just integrate the whole thing normally..the f(x) and f'(x) parts were putting me off
Yeah, that's just showing you what you plug into the integral

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