When a curve is rotated around the x-axis, it produces a three-dimensional surface. Calculating the area of this surface may initially seem complicated, especially when the curve is given in parametric form.

The key is to understand where the surface-area formula comes from rather than simply memorizing it.

Consider the parametric curve

x = x(t)

y = y(t)

When this curve is rotated about the x-axis, the surface area is given by

$S=2\pi\int y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

Let’s derive this formula first and then apply it to a specific example.

1. What Happens When a Curve Is Rotated?

Imagine a small piece of the curve between two very close points, A and B.

Suppose point A is approximately at

(x, y)

and point B is slightly farther along the curve.

The distance between A and B along the curve is called ds.

Now rotate this tiny piece of the curve around the x-axis.

The result is a very small strip of three-dimensional surface.

The easiest way to understand this strip is to imagine a tiny frustum of a cone.

Its circumference is approximately

$2\pi y$

because the distance from the curve to the x-axis is y.

Its slanted width is approximately ds.

Therefore, its tiny surface area is

$dS=2\pi y,ds$

This simple relationship is the foundation of the surface-area formula.

2. Why Is the Circumference 2πy?

When the point (x, y) rotates around the x-axis, it travels in a circle.

The radius of that circle is the distance from the point to the x-axis.

That distance is y.

The circumference of a circle is

$\text{circumference}=2\pi\times\text{radius}$

Therefore,

$\text{circumference}=2\pi y$

So the tiny surface strip has approximately

$dS=\text{circumference}\times\text{width}$

which gives

$dS=2\pi y,ds$

3. What Exactly Is ds?

The symbol ds means a tiny piece of arc length along the curve.

This is important.

It is not simply the horizontal distance dx.

It is not simply the vertical distance dy.

The curve may be sloping, so its actual length is generally different from either dx or dy.

Consider a tiny piece of the curve.

The horizontal change is dx.

The vertical change is dy.

These form the two sides of a tiny right triangle.

The actual distance along the curve is approximately the hypotenuse.

Therefore, by the Pythagorean theorem,

$ds^2=dx^2+dy^2$

Taking the square root gives

$ds=\sqrt{dx^2+dy^2}$

This is the basic geometric idea behind the arc-length element.

4. Converting ds for a Parametric Curve

For an ordinary curve y = f(x), we can work directly with x.

But a parametric curve is different.

Both x and y depend on another variable, t:

$x=x(t)$ $y=y(t)$

Therefore, changes in x and y can be expressed in terms of changes in t.

We have

$dx=\frac{dx}{dt},dt$

and

$dy=\frac{dy}{dt},dt$

Substitute these into

$ds=\sqrt{dx^2+dy^2}$

We get

$ds=\sqrt{\left(\frac{dx}{dt},dt\right)^2+\left(\frac{dy}{dt},dt\right)^2}$

For increasing t, dt is positive, so we can take dt outside the square root:

$ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

This is the arc-length element for a parametric curve.

5. Put This into the Surface-Area Formula

We already established that

$dS=2\pi y,ds$

And we just found

$ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

Substituting this expression for ds gives

$dS=2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

To find the entire surface area, we add all the tiny surface pieces together.

Adding infinitely many tiny pieces is what integration does.

Therefore,

$S=2\pi\int y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

This is the surface-area formula for a parametric curve rotated about the x-axis when y is the distance from the curve to the axis of rotation.

6. Understanding the Formula Intuitively

The formula

$S=2\pi\int y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

can be understood as:

Surface area = circumference × arc length

More specifically,

$2\pi y$

is the circumference of the circle created by rotation.

And

$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

is the tiny arc length ds.

Therefore,

$dS=2\pi y,ds$

The integral simply adds all these tiny surface strips together.

7. Apply the Formula to Our Problem

Now consider the curve

$x=3\cos t+\cos 3t$ $y=3\sin t-\sin 3t$

with

$0\leq t\leq2\pi$

The curve is rotated about the x-axis.

Our formula is

$S=2\pi\int y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

So we need to find three things:

y(t)
dx/dt
dy/dt

Let’s calculate each one.

8. Simplifying y(t)

We start with

$y=3\sin t-\sin 3t$

Recall the triple-angle identity

$\sin 3t=3\sin t-4\sin^3t$

Substitute this:

$y=3\sin t-(3\sin t-4\sin^3t)$

The 3 sin t terms cancel:

$y=4\sin^3t$

$y(t)=4\sin^3t$

This tells us the distance of the curve from the x-axis at each value of t on the upper half of the curve.

9. Finding dx/dt

We have

$x=3\cos t+\cos 3t$

Differentiate:

$\frac{dx}{dt}=-3\sin t-3\sin 3t$

Factor out -3:

$\frac{dx}{dt}=-3(\sin t+\sin 3t)$

Using the identity

$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

we get

$\sin t+\sin 3t=2\sin 2t\cos t$

Therefore,

$\frac{dx}{dt}=-6\sin 2t\cos t$

10. Finding dy/dt

Starting with

$y=3\sin t-\sin 3t$

Differentiate:

$\frac{dy}{dt}=3\cos t-3\cos 3t$

Factor:

$\frac{dy}{dt}=3(\cos t-\cos 3t)$

Using the identity

$\cos A-\cos B=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

we get

$\cos t-\cos 3t=2\sin 2t\sin t$

Therefore,

$\frac{dy}{dt}=6\sin 2t\sin t$

11. Calculate the Arc-Length Factor

The formula requires

$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$

We have

$\frac{dx}{dt}=-6\sin 2t\cos t$

and

$\frac{dy}{dt}=6\sin 2t\sin t$

Therefore,

$\sqrt{(-6\sin 2t\cos t)^2+(6\sin 2t\sin t)^2}$

Squaring gives

$\sqrt{36\sin^2 2t\cos^2t+36\sin^2 2t\sin^2t}$

Factor:

$\sqrt{36\sin^2 2t(\cos^2t+\sin^2t)}$

Since

$\cos^2t+\sin^2t=1$

this becomes

$\sqrt{36\sin^2 2t}$

Therefore,

$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=6|\sin 2t|$

The absolute value is necessary because the square root always gives a non-negative result.

12. Why Don’t We Use 0 to 2π?

This is another important idea.

The original parametric equations are given for

$0\leq t\leq2\pi$

This traces the entire astroid.

However, we are rotating the curve around the x-axis.

The portion from

$0\leq t\leq\pi$

is the upper half of the curve.

When this upper half is rotated around the x-axis, it generates the entire three-dimensional surface.

The lower half, corresponding to

$\pi\leq t\leq2\pi$

would generate the same surface again.

So if we integrated from 0 to 2π while using the distance from the x-axis as the radius, we would count the same surface twice.

Therefore, we only need

$0\leq t\leq\pi$

13. Substitute Everything into the Formula

We have

$y=4\sin^3t$

and

$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=6|\sin 2t|$

The formula is

$S=2\pi\int y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

Therefore,

$S=2\pi\int_0^\pi(4\sin^3t)(6|\sin 2t|),dt$

Multiply the constants:

$S=48\pi\int_0^\pi\sin^3t|\sin 2t|,dt$

Now use

$\sin 2t=2\sin t\cos t$

Therefore,

$|\sin 2t|=2\sin t|\cos t|$

on the interval 0 to π.

So,

$S=48\pi\int_0^\pi\sin^3t(2\sin t|\cos t|),dt$

Finally,

$S=96\pi\int_0^\pi\sin^4t|\cos t|,dt$

Therefore, the expression required for the integral is

$96\pi\sin^4(t)|\cos(t)|$

14. Evaluate the Surface Area

We can go one step further and actually calculate the surface area.

We have

$S=96\pi\int_0^\pi\sin^4t|\cos t|,dt$

The integrand is symmetric around

$t=\frac{\pi}{2}$

Therefore,

$S=192\pi\int_0^{\pi/2}\sin^4t\cos t,dt$

Let

$u=\sin t$

Then

$du=\cos t,dt$

When t = 0,

$u=0$

When t = π/2,

$u=1$

Therefore,

$S=192\pi\int_0^1u^4,du$

Integrating:

$S=192\pi\left[\frac{u^5}{5}\right]_0^1$

So,

$S=\frac{192\pi}{5}$

Final Result

The surface-area integral is

$S=96\pi\int_0^\pi\sin^4t|\cos t|,dt$

The resulting surface area is

$S=\frac{192\pi}{5}$

The Big Picture

The entire derivation can be summarized in one chain of ideas:

Tiny arc of curve → ds

Rotate it around the x-axis → tiny surface strip

Tiny surface strip = circumference × arc length

$dS=2\pi y,ds$

For a parametric curve,

$ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

Therefore,

$dS=2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

And adding all the strips gives

$S=2\pi\int y(t)\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt$

So the formula is not something to memorize blindly.

It comes directly from the simple geometric idea:

Surface area = circumference × tiny slanted length.

Finding the Surface Area of a Parametric Curve Rotated About the x-Axis

1. What Happens When a Curve Is Rotated?

2. Why Is the Circumference 2πy?

3. What Exactly Is ds?

4. Converting ds for a Parametric Curve

5. Put This into the Surface-Area Formula

6. Understanding the Formula Intuitively

7. Apply the Formula to Our Problem

8. Simplifying y(t)

9. Finding dx/dt

10. Finding dy/dt

11. Calculate the Arc-Length Factor

12. Why Don’t We Use 0 to 2π?

13. Substitute Everything into the Formula

14. Evaluate the Surface Area

Final Result

The Big Picture

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DavidsonNext: AP® Calculus: Challenging Concepts from Calculus AB & Calculus BC

Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

1. What Happens When a Curve Is Rotated?

2. Why Is the Circumference 2πy?

3. What Exactly Is ds?

4. Converting ds for a Parametric Curve

5. Put This into the Surface-Area Formula

6. Understanding the Formula Intuitively

7. Apply the Formula to Our Problem

8. Simplifying y(t)

9. Finding dx/dt

10. Finding dy/dt

11. Calculate the Arc-Length Factor

12. Why Don’t We Use 0 to 2π?

13. Substitute Everything into the Formula

14. Evaluate the Surface Area

Final Result

The Big Picture

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Like this:

DavidsonNext: AP® Calculus: Challenging Concepts from Calculus AB & Calculus BC

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series