## 17Calculus Parametric Equations - Surface Area

##### 17Calculus

Calculate the area of a surface that is described by parametric equations.

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. $S = 2\pi ~ \int_{t_0}^{t_1}{ Y(t) \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~ dt}$ If you compare this integral to the equation for arc length (on the prevous page) 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 $S = 2\pi ~ \int_{t_0}^{t_1}{Y(t) ~ ds}$ where $$ds = \sqrt{[X'(t)]^2 + [Y'(t)]^2} ~dt$$

When rotating about the y-axis, the integral we use is $S = 2\pi ~ \int_{t_0}^{t_1}{X(t) ~ ds}$ 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.

See the resources section for links to pages explaining the derivation of these equations.

Practice

Unless otherwise instructed, calculate the surface area formed by revolving the parametric curve about the given axis.

$$x = t^4+4$$, $$y = 8t$$, $$0 \leq t \leq 2$$; $$x$$-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

In this video, when he sets up his integral, he initially places the $$Y(t)$$ term outside the integral. Then he moves the $$t$$ inside the integral to integrate. These steps are not correct. The $$Y(t)$$ term should be inside the integral the entire time. He does eventually stumble into the correct answer but his procedure is not correct.

### The Organic Chemistry Tutor - 3507 video solution

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$$x = t^2$$, $$y = t^3$$, $$0 \leq t \leq 1$$; $$y$$-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

In this video, when he sets up his integral, he initially places the $$X(t)$$ term outside the integral. Then he moves the $$t$$ inside the integral to integrate. These steps are not correct. The $$X(t)$$ term should be inside the integral the entire time. He does eventually stumble into the correct answer but his procedure is not correct.

### The Organic Chemistry Tutor - 3508 video solution

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$$x = 3t^2$$, $$y = 2t^3$$, $$0 \leq t \leq 5$$; $$y$$-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.

Solution

### Krista King Math - 594 video solution

video by Krista King Math

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$$x = 1-t$$, $$y = 2\sqrt{t}$$, $$1 \leq t \leq 4$$; $$x$$-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

### Krista King Math - 59 video solution

video by Krista King Math

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$$x = 3 + 2t$$, $$y = 9 - 3t$$, $$1 \leq t \leq 4$$; $$y$$-axis

Problem Statement

Find the surface area of revolution of the parametric curve $$x = 3 + 2t$$, $$y = 9 - 3t$$, $$1 \leq t \leq 4$$ rotated about the $$y$$-axis.

Solution

The solution can be found on this page.

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$$x = 9 + 2t^2$$, $$y = 4t$$, $$0 \leq t \leq 2$$; $$x$$-axis

Problem Statement

Find the surface area of revolution of the parametric curve $$x = 9 + 2t^2$$, $$y = 4t$$, $$0 \leq t \leq 2$$ rotated about the $$x$$-axis.

Solution

The solution can be found on this page.

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$$x = 3\cos(\pi t)$$, $$y = 5t + 2$$, $$0 \leq t \leq 1/2$$; $$y$$-axis

Problem Statement

Find the surface area of revolution of the parametric curve $$x = 3\cos(\pi t)$$, $$y = 5t + 2$$, $$0 \leq t \leq 1/2$$ rotated about the $$y$$-axis.

Solution

The solution can be found on this page.

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Set up, but do not evaluate, an integral that calculates the surface area of revolution of the curve $$x = 1 + \ln(5 + t^2)$$, $$y = 2t - 2t^2$$, $$0 \leq t \leq 2$$ rotated about the $$x$$-axis.

Problem Statement

Set up, but do not evaluate, an integral that calculates the surface area of revolution of the curve $$x = 1 + \ln(5 + t^2)$$, $$y = 2t - 2t^2$$, $$0 \leq t \leq 2$$ rotated about the $$x$$-axis.

Solution

The solution can be found on this page.

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Set up, but do not evaluate, an integral that calculates the surface area of revolution of the curve $$x = 1 + 3t^2$$, $$y = \sin(2t)\cos(t/4)$$, $$0 \leq t \leq 1/2$$ rotated about the $$y$$-axis.

Problem Statement

Set up, but do not evaluate, an integral that calculates the surface area of revolution of the curve $$x = 1 + 3t^2$$, $$y = \sin(2t)\cos(t/4)$$, $$0 \leq t \leq 1/2$$ rotated about the $$y$$-axis.

Solution

The solution can be found on this page.

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