## 17Calculus Parametric Equations - Surface Area

On this page we discuss integration and surface area of parametric equations.

To calculate the surface area defined by revolving a parametric curve defined as $$x=X(t)$$ and $$y=Y(t)$$ from $$t=t_0$$ to $$t=t_1$$ about the x-axis, we use this integral.

$$\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{ Y(t) \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~ dt}}$$

If you compare this integral to the equation for arc length (in the prevous section) you will see the common factor $$\sqrt{[X'(t)]^2 + [Y'(t)]^2}$$

Some textbooks write the surface integral differently, taking this into account. You may see it written as

$$\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{Y(t) ~ ds}}$$
where $$ds = \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~dt$$

When rotating about the y-axis, the integral we use is

$$\displaystyle{S = 2\pi ~ \int_{t_0}^{t_1}{X(t) ~ ds}}$$

Notice we use a capital S to represent surface area. This is the standard symbol you will see in many textbooks. We reserve a lowercase s to represent arc length.

Practice

Find the surface area of revolution of the parametric curve $$x = t^4+4$$, $$y = 8t$$, $$0 \leq t \leq 2$$ rotated about the x-axis.

Problem Statement

Find the surface area of revolution of the parametric curve $$x = t^4+4$$, $$y = 8t$$, $$0 \leq t \leq 2$$ rotated about the x-axis.

Solution

### 3507 video

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Find the surface area of revolution of the parametric curve $$x = t^2$$, $$y = t^3$$, $$0 \leq t \leq 1$$ rotated about the y-axis.

Problem Statement

Find the surface area of revolution of the parametric curve $$x = t^2$$, $$y = t^3$$, $$0 \leq t \leq 1$$ rotated about the y-axis.

Solution

### 3508 video

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Find the surface area of revolution of the parametric curve $$x = 3t^2$$, $$y = 2t^3$$, $$0 \leq t \leq 5$$ rotated about the y-axis.

Problem Statement

Find the surface area of revolution of the parametric curve $$x = 3t^2$$, $$y = 2t^3$$, $$0 \leq t \leq 5$$ rotated about the y-axis.

$$\displaystyle{ S_y = \frac{24\pi}{5} \left[ 949\sqrt{26}+1 \right] }$$

Problem Statement

Find the surface area of revolution of the parametric curve $$x = 3t^2$$, $$y = 2t^3$$, $$0 \leq t \leq 5$$ rotated about the y-axis.

Solution

### 594 video

video by Krista King Math

$$\displaystyle{ S_y = \frac{24\pi}{5} \left[ 949\sqrt{26}+1 \right] }$$

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Find the surface area of revolution of the parametric curve $$x = 1-t$$, $$y = 2\sqrt{t}$$, $$1 \leq t \leq 4$$ rotated about the x-axis.

Problem Statement

Find the surface area of revolution of the parametric curve $$x = 1-t$$, $$y = 2\sqrt{t}$$, $$1 \leq t \leq 4$$ rotated about the x-axis.

Solution

### 59 video

video by Krista King Math

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### Related Topics and Links

 Wikipedia - Parametric Derivative Wikipedia - Arc Length Pauls Online Notes - Arc Length with Parametric Equations Wikipedia - Solid of Revolution

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

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

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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