Calculating the area of a curve rotating around an axis

**Sabo** · April 7th 2011, 06:47 AM

Calculate the area of the rotational surface that arise when the curve $y=^\frac{x^2}{8}, \, 0\leqslant y\leqslant 6$ rotates around the y-axis.

My approach is something along the lines of:
Use an integral to calculate the area.
Since the curve is rotating around the y-axis, we need to (not sure about the english term), "extract" x from the formula (we need to get the height between the curve and the y-axis).

$y=\frac{x^2}{8}\Leftrightarrow \frac{y}{8}=x^2 \Rightarrow x=\pm\sqrt{8y}$

Since the curve is symmetrical around the y-axis we don't have to care the negative x; we simply multiply by 2 to get the total area.

So, we now have the following

$\int_0^6 \sqrt{8y} \, dy$

which simplified should be

$2\sqrt{2}\int_0^6 \sqrt{y} \, dy = 2\sqrt{2}[\frac{2\sqrt[3]{6}}{3}]$

Something isn't right though, and I can't figure out what.

**NOX Andrew** · April 7th 2011, 06:49 AM

You found the area of the region, not the surface area of the surface of revolution. Have you learned the formula for the surface area of a surface of revolution? If not, then here is an article about surfaces of revolution.

The formula for the surface area is

$\displaystyle S = 2\pi \int_c^d f(y) \sqrt{1 + \left[f'(x)\right]^2} \, dy$

**Sabo** · April 7th 2011, 08:55 AM

Ah, big failure on my part. I was stuck in 2D mode. The surface area is in the next sub-chapter the book. I apologize for taking up your time. =/

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