Finding surface of revolution

Question

If a curve in the $y-z$ plane given by $z=f(y)$ is rotated around the y-axis, what will be the surface of revolution?

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How should I approach this problem? Any help will be appreciated.

Edit: I think the answers below are for the surface area, which is nice but actually I want to find the equation of the surface.

Answer 1

Observe that the intersection between the revolved surface and a plane normal to the $y$-axis and at any given $y$ is a circle of radius $f(y)$ in the $xz$-coordinates.

Thus, the equation of the surface as a result of $f(y)$ revolving around the $y$-axis is given by,

$$x^2+z^2 =[f(y)]^2$$

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Think of the whole surface as a collection of rings of width $ds$, which is just the length of the curve $z(y)$ within $dy$ given by

$$ds=\sqrt{1+[z(y)’]^2}dy$$

Then, the surface area for each ring is $2\pi zds$ and the surface integral is

$$\int_{y_1}^{y_2} 2\pi z ds=2\pi \int_{y_1}^{y_2} z(y)\sqrt{1+[z(y)’]^2}dy$$.

Answer 2

The surface element is given by

$$dS=2\pi f(y)dH=2\pi f(y)\sqrt{dy^2+dz^2}=2\pi f(y)\sqrt{1+[f'(y)]^2}dy$$

then integrate from $y_{min}$ to $y_{max}$.

Refer also to Surface of revolution.

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Answer 3

The formula is $$ S= \int _a ^b 2\pi f(y)\sqrt {1+(f'(y))^2}dy$$

The integral is with respect to $y$

Answer 4

Direction of vector $x$ axis is the same in either case.. coming out of paper.

Normally used $(y, z)$ are swapped to become $ (z,y)$

1) If $y$ axis is axis of revolution

$$ z=g(y)$$

$$ Surface Area= 2 \pi \int z \sqrt{1+(dz/dy)^2} dy $$

evaluated between $y$ limits of integration.

2) If $z$ axis is axis of revolution

$$ y=f(z)$$

$$ Surface Area= 2 \pi \int y \sqrt{1+(dy/dz)^2} dz $$

evaluated between $z$ limits

Answer 5

$$z = f \left( \sqrt{x^2+y^2} \right)$$