## 17Calculus Integrals - Surface Area

### Integrals

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Length-Area-Volume

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This page covers the topic of surface area of an explicitly defined smooth curve revolved around an axis in the xy-plane in cartesian (rectangular) coordinates. [You can also calculate surface area in polar coordinates and for surfaces described parametrically.]

Setting up the integral to calculate surface area is fairly straight-forward. The difficulty with this topic occurs when evaluating the integral, which can quickly become quite complicated. Consequently, most problems you get will be carefully hand-picked by your instructor or the textbook author so that you can evaluate the integrals with the techniques you know. The comments we made on the arc length page about tricks to evaluating these integrals apply here as well.

First, let's look at a video clip explaining how to derive the surface area equations.

### MIP4U - Surface Area of Revolution - Part 1 of 2 [5min-16secs]

video by MIP4U

$$\int_{a}^{b}{2\pi y ~ ds}$$

$$\int_{c}^{d}{2\pi x ~ ds}$$

$$ds = \sqrt{1 + [f'(x)]^2} ~dx$$ or $$ds = \sqrt{1 + [g'(y)]^2} ~dy$$

Notes
1. $$ds$$ in the last row above is the differential arc length as discussed on the arc length page. Using $$ds$$ allows us to write the integral in a more compact form and it is easier to see what is going on.
2. Which $$ds$$ you use depends on how the graph is described.

Here is a great video clip explaining these equations in more detail.

### PatrickJMT - Finding Surface Area - Part 1 [1min-2secs]

video by PatrickJMT

Practice

Unless otherwise instructed, calculate the surface area of the given line segment rotated about the given axis. Give all answers in exact form.

 equation: $$y=\sqrt{x}$$ range: $$4\leq x\leq9$$ axis of rotation: x-axis

Problem Statement

Calculate the surface area of the line segment $$y=\sqrt{x}$$; $$4\leq x\leq9$$ rotated about the x-axis. Give your answer in exact form.

Solution

He works this problem twice in two videos, using different $$ds$$ equations.

### 1195 video

video by PatrickJMT

### 1195 video

video by PatrickJMT

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 equation: $$y=x^2-\frac{1}{8}\ln(x)$$ range: $$1\leq x\leq2$$ axis of rotation: y-axis

Problem Statement

Calculate the surface area of the line segment $$y=x^2-\frac{1}{8}\ln(x)$$; $$1\leq x\leq2$$ rotated about the y-axis. Give your answer in exact form.

Solution

### 1196 video

video by Krista King Math

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 equation: $$y=\sqrt{4-x^2}$$ range: $$-1\leq x\leq1$$ axis of rotation: x-axis

Problem Statement

Calculate the surface area of the line segment $$y=\sqrt{4-x^2}$$; $$-1\leq x\leq1$$ rotated about the x-axis. Give your answer in exact form.

Solution

### 1197 video

video by Krista King Math

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 equation: $$f(x)=(1/3)x^3$$ range: $$[0,2]$$ axis of rotation: x-axis

Problem Statement

Calculate the surface area of the line segment $$f(x)=(1/3)x^3$$; $$[0,2]$$ rotated about the x-axis. Give your answer in exact form.

Solution

### 1198 video

video by MIP4U

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 equation: $$f(x)=\sqrt[3]{x}$$ range: $$[0,8]$$ axis of rotation: y-axis

Problem Statement

Calculate the surface area of the line segment $$f(x)=\sqrt[3]{x}$$; $$[0,8]$$ rotated about the y-axis. Give your answer in exact form.

Solution

### 1199 video

video by MIP4U

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 equation: $$y=\sqrt[3]{6x}$$ range: $$0 \leq x \leq 4/3$$ axis of rotation: y-axis

Problem Statement

Calculate the surface area of the line segment $$y=\sqrt[3]{6x}$$; $$0 \leq x \leq 4/3$$ rotated about the y-axis. Give your answer in exact form.

Solution

### 1915 video

video by MIP4U

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 equation: $$f(x)=\sqrt{x}$$ range: $$0\leq x\leq 4$$ axis of rotation: x-axis

Problem Statement

Calculate the surface area of the line segment $$f(x)=\sqrt{x}$$; $$0\leq x\leq 4$$ rotated about the x-axis. Give your answer in exact form.

$$[(17)^{3/2}-1]\pi/6$$

Problem Statement

Calculate the surface area of the line segment $$f(x)=\sqrt{x}$$; $$0\leq x\leq 4$$ rotated about the x-axis. Give your answer in exact form.

Solution

### 2002 video

video by Dr Chris Tisdell

$$[(17)^{3/2}-1]\pi/6$$

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You CAN Ace Calculus

 integration arc length

### 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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