MHB Adwt's question at Yahoo Answers regarding surfaces of revolution

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The discussion focuses on calculating the surface area of the curve defined by y=4+3x^2, revolving around the y-axis between x=1 and x=2. The appropriate formula for surface area is provided, which involves integrating a specific function derived from the curve's derivative. A u-substitution is applied to simplify the integral, leading to the computation of the surface area. The final result of the surface area calculation is approximately 88.49. This method demonstrates the application of calculus techniques to solve real-world geometric problems.
MarkFL
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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.
 
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Hello adwt,

The formula we want to use is:

$$S=2\pi\int_a^b x\sqrt{1+\left(f'(x) \right)^2}\,dx$$

We are given:

$$a=1,\,b=2,\,f(x)=4+3x^2\,\therefore\,f'(x)=6x$$

Hence, we have:

$$S=2\pi\int_1^2 x\sqrt{1+36x^2}\,dx$$

Let's use a $u$-subsitution:

$$u=1+36x^2\,\therefore\,du=72x\,dx$$

And we may now write:

$$S=\frac{\pi}{36}\int_{37}^{145}u^{\frac{1}{2}}\,du$$

Applying the FTOC, along with the power rule for integration we find:

$$S=\frac{\pi}{54}\left[u^{\frac{3}{2}} \right]_{37}^{145}=\frac{\pi}{54}\left(145\sqrt{145}-37\sqrt{37} \right)\approx88.4863895868960$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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