Surface area of revolution for curves symmetrical on the axis of revolution.

Question

I understand that surface area generated when an curve is rotated on the x axis by $2\pi$ radians is given by: $2\pi∫yds$

How is this area affected when the object is symmetrical on the x-axis, e.g. an ellipse. Would you double the area given by the equation if it is rotated by $2\pi$? If you were to rotate it by just $\pi$, would you still use the equation for $2\pi$?

Thanks.

Answer 1

I would just find the area generated when the top half of the curve is rotated about the $x$-axis. So for example if the curve is the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a$ and $b$ are positive, find the area generated when we rotate the curve $y=b\sqrt{1-\frac{x^2}{a^2}}$ about the $x$-axis.

Answer 2

To answer the last part of your question you can see that if you rotate the curve only $\pi$, you have to multiply by 2 anyways to give an equivalent surface area.