Area of surface of revolution

Question

Write an integral with respect to a that represents the area of the surface obtained by rotating the curve

y = x³, -1 < to < 1 about the line x = 3. Compute any required derivatives. You do not need to evaluate

the integral. Then Write an integral for the same quantity with respect to y.

Answer 1

Let's take a detoured approach to solving this that will help you understand geometrically what's going on.

Recall that if a function ƒ(x) is rotated around the y-axis, the area of the surface of revolution is given by

A= 2 π∫ x ds

ds=(1+ [f'(x)]^2 )^1/2 dx

where the integral is over the desired positive domain [a,b] for which f(x) is rotated and ds is the differential line element.

We can reduce the problem of rotating y=x³ about the line x=3 to the above case as follows. Note that surface area is preserved under rigid translations and reflections in space.

So we can equivalently rotate the function

f(x) = (x+3)³

about the y-axis. You should understand why this is equivalent.

x³ —>(x+3)³ is a horizontal translation to the left by three. Draw a picture to see that rotating f(x) around y-axis gives your desired surface translated by three to the left.

Now setting up the integral, its easiest to take the function

f(x) = -x(x-3)³

on the positive domain [2,4]. Now you should be asking why this function? Well, if you plot this you'll notice that it is simply the reflection of (x+3)³ across the y-axis, but we can examine it directly.

x³ —> -x³—> -(x-3)³

where the first arrow denotes reflection about the y-axis and the second arrow denotes translation to the right by 3. The surface is the same!

This reduces to rotating a function defined on a positive domain about the y-axis. So we can use the general formula above.

A= 2 π∫ x ds

ds=(1+ [f'(x)]^2 )^1/2 dx

f(x) = -(x-3)³ –> f'(x) = -3(x-3)^{^2} for x in [2,4]

Answer 2

Generally rotating a function around a vertical will be π*(integral) of rds. where r is the horizontal distance from the function to the vertical axis and ds is the differential of arc length.

We will put the whole integral in terms of y:

r(y) = 3-x(y) = 3 - y^1/3

ds = sqrt((dx/dy)² + 1)dy = sqrt(((1/3)y^-2/3)² + 1) dy

y limits are -1 to 1

So the integral is integral from -1 to 1 of π(3-y^1/3)sqrt(1/(9y^4/3) + 1) dy