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adwt's question at Yahoo! Answers regarding surfaces of revolution
- Thread starter MarkFL
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Hello adwt,
The formula we want to use is:
\(\displaystyle S=2\pi\int_a^b x\sqrt{1+\left(f'(x) \right)^2}\,dx\)
We are given:
\(\displaystyle a=1,\,b=2,\,f(x)=4+3x^2\,\therefore\,f'(x)=6x\)
Hence, we have:
\(\displaystyle S=2\pi\int_1^2 x\sqrt{1+36x^2}\,dx\)
Let's use a $u$-subsitution:
\(\displaystyle u=1+36x^2\,\therefore\,du=72x\,dx\)
And we may now write:
\(\displaystyle S=\frac{\pi}{36}\int_{37}^{145}u^{\frac{1}{2}}\,du\)
Applying the FTOC, along with the power rule for integration we find:
\(\displaystyle S=\frac{\pi}{54}\left[u^{\frac{3}{2}} \right]_{37}^{145}=\frac{\pi}{54}\left(145\sqrt{145}-37\sqrt{37} \right)\approx88.4863895868960\)
The formula we want to use is:
\(\displaystyle S=2\pi\int_a^b x\sqrt{1+\left(f'(x) \right)^2}\,dx\)
We are given:
\(\displaystyle a=1,\,b=2,\,f(x)=4+3x^2\,\therefore\,f'(x)=6x\)
Hence, we have:
\(\displaystyle S=2\pi\int_1^2 x\sqrt{1+36x^2}\,dx\)
Let's use a $u$-subsitution:
\(\displaystyle u=1+36x^2\,\therefore\,du=72x\,dx\)
And we may now write:
\(\displaystyle S=\frac{\pi}{36}\int_{37}^{145}u^{\frac{1}{2}}\,du\)
Applying the FTOC, along with the power rule for integration we find:
\(\displaystyle S=\frac{\pi}{54}\left[u^{\frac{3}{2}} \right]_{37}^{145}=\frac{\pi}{54}\left(145\sqrt{145}-37\sqrt{37} \right)\approx88.4863895868960\)