Adwt's question at Yahoo Answers regarding surfaces of revolution

[MarkFL]

Here is the question:

Calculus - Surface Area of y=4+3x^2?


What is the surface area of y=4+3x^2 from where x = [1,2] about the y-axis?

Please include work/explanation.

Thanks.

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

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:

\(\displaystyle S=\frac{\pi}{54}\left[u^{\frac{3}{2}} \right]_{37}^{145}=\frac{\pi}{54}\left(145\sqrt{145}-37\sqrt{37} \right)\approx88.4863895868960\)