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#### MarkFL

Staff member
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?

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

#### MarkFL

Staff member

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:

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