On this page we discuss integration and surface area of parametric equations.
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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 xaxis, 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 yaxis, 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 xaxis.
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 xaxis.
Solution 

video by The Organic Chemistry Tutor 

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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 yaxis.
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 yaxis.
Solution 

video by The Organic Chemistry Tutor 

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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 yaxis.
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 yaxis.
Final Answer 

\(\displaystyle{ S_y = \frac{24\pi}{5} \left[ 949\sqrt{26}+1 \right] }\)
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 yaxis.
Solution 

video by Krista King Math 

Final Answer 

\(\displaystyle{ S_y = \frac{24\pi}{5} \left[ 949\sqrt{26}+1 \right] }\)
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Find the surface area of revolution of the parametric curve \(x = 1t\), \(y = 2\sqrt{t}\), \(1 \leq t \leq 4\) rotated about the xaxis.
Problem Statement 

Find the surface area of revolution of the parametric curve \(x = 1t\), \(y = 2\sqrt{t}\), \(1 \leq t \leq 4\) rotated about the xaxis.
Solution 

video by Krista King Math 

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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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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