Area of a Surface of Revolution

A surface of revolution is obtained when a curve is rotated about an axis.

We consider two cases - revolving about the x-axis and revolving about the y-axis.

Revolving about the x-axis

Suppose that y (x), y (t), and y (θ) are smooth non-negative functions on the given interval.

1. If the curve y = f (x), a ≤ x ≤ b is rotated about the x-axis, then the surface area is given by
   $$A=2\pi \underset{a}{\overset{b}{\int }}f\left(x\right)\sqrt{1+{\left[f^{\prime }\left(x\right)\right]}^2}dx.$$

   

2. If the curve is described by the function $x=g\left(y\right),$ $c\le y\le d,$ and rotated about the $x-$axis, then the area of the surface of revolution is given by
   $$A=2\pi \underset{c}{\overset{d}{\int }}y\sqrt{1+{\left[g^{\prime }\left(y\right)\right]}^2}dy.$$

   

3. If the curve defined by the parametric equations $x=x\left(t\right),$ $y=y\left(t\right),$ with $t$ ranging over some interval $\left[\alpha ,\beta \right],$ is rotated about the $x-$axis, then the surface area is given by the following integral (provided that $y\left(t\right)$ is never negative)
   $$A=2\pi \underset{\alpha }{\overset{\beta }{\int }}y\left(t\right)\sqrt{{\left[x^{\prime }\left(t\right)\right]}^2+{\left[y^{\prime }\left(t\right)\right]}^2}dt.$$

   

4. If the curve defined by polar equation $r=r\left(\theta \right),$ with $\theta$ ranging over some interval $\left[\alpha ,\beta \right],$ is rotated about the polar axis, then the area of the resulting surface is given by the following formula (provided that $y=r\sin \theta$ is never negative)
   $$A=2\pi \underset{\alpha }{\overset{\beta }{\int }}r\left(\theta \right)\sin \theta \sqrt{{\left[r\left(\theta \right)\right]}^2+{\left[r^{\prime }\left(\theta \right)\right]}^2}d\theta .$$

   

Revolving about the $y-$axis

The functions $$ $$ and $$ are supposed to be smooth and non-negative on the given interval.

1. If the curve $$ $$ is rotated about the $$axis, then the surface area is given by
   $$

   

2. If the curve is described by the function $$ $$ and rotated about the $$axis, then the area of the surface of revolution is given by
   $$

   

3. If the curve defined by the parametric equations $$ $$ with $$ ranging over some interval $$ is rotated about the $$axis, then the surface area is given by the integral (provided that $$ is never negative)
   $$

   

4. If the curve defined by polar equation $$ with $$ ranging over some interval $$ is rotated about the $$axis, then the area of the resulting surface is given by the formula (provided that $$ is never negative)
   $$

   

Solved Problems

Solution.

Let the slant height $$ be defined by the equation $$ The slope $$ is given by

where $$ is the height of the cone.

We calculate the lateral surface area of the cone by the formula

Substituting

we obtain

By the Pythagorean theorem, $$ Hence,

Solution.

We find the surface area through integration by the formula

We integrate here from $$ to $$ As $$ then

So we have

Recall the following hyperbolic identities:

This yields:

Solution.

When calculating the surface area, we consider the part of the astroid lying in the first quadrant and then multiply the result by $$ As the curve is defined in parametric form, we can write

Find the derivatives:

and simplify the radicand:

Hence, the surface area is

Solution.

We determine the surface area by the formula

Due to symmetry, we can integrate from $$ to $$ considering the curve in the first quadrant and then multiply the result by $$ So

Take the derivative:

Hence, the derivative squared is written in the form

We can simplify the expression with the square root:

Then