\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Parametric Equations - Area

Limits

Using Limits

Limits FAQs

Derivatives

Graphing

Related Rates

Optimization

Other Applications

Integrals

Improper Integrals

Trig Integrals

Length-Area-Volume

Applications - Tools

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Tools

Calculus Tools

Additional Tools

Articles

Parametrics

SV Calculus

MV Calculus

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Tools

Calculus Tools

Additional Tools

Articles

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

The area under a smooth curve defined parametrically as \( x = X(t) \) and \( y = Y(t) \) from \( t=t_0\) to \( t=t_1\) can be calculated using the integral

\(\displaystyle{s = \int_{t_0}^{t_1}{Y(t)X'(t) ~ dt}}\)     where \( X'(t) = dX/dt \).

You may also find this written in a shorthand form as

\(\displaystyle{s = \int_{t_0}^{t_1}{y~dx}}\)

In this notation, \( dx = (dx/dt) dt \).

Practice

Calculate the area under the parametric curve \( x = t^2 \), \( y = 4t^2-t^4 \) for \( 0 \leq t \leq 2 \).

Problem Statement

Calculate the area under the parametric curve \( x = t^2 \), \( y = 4t^2-t^4 \) for \( 0 \leq t \leq 2 \).

Solution

3504 video

close solution

Log in to rate this practice problem and to see it's current rating.

Find the area under the curve \(x=1+e^t\), \(y=t-t^2\) and above the x-axis.

Problem Statement

Find the area under the curve \(x=1+e^t\), \(y=t-t^2\) and above the x-axis.

Solution

57 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

Calculate the area enclosed by the parametric curve \(x = t^3 - 3t\), \(y = 4t^2\).

Problem Statement

Calculate the area enclosed by the parametric curve \(x = t^3 - 3t\), \(y = 4t^2\).

Final Answer

Area = \(\displaystyle{ \frac{96}{5}\sqrt{3} }\)

Problem Statement

Calculate the area enclosed by the parametric curve \(x = t^3 - 3t\), \(y = 4t^2\).

Solution

The most difficult and time-consuming part of this problem is setting up the integral. Here is a general overview explaining how we will tackle this problem.
1. Graph the parametric equation.
2. Find the important points that determine the area to be calculated.
3. Determine how to set up the integrals in general to calculate the area.
4. Set up the integrals and evaluate.

1. Graph the parametric equation.
Notice we have included the direction of the graph in the plot. Now we can easily see the area that needs to be calculated. Without this graph, it would have been extremely difficult to set up the integral(s) to determine the area.

2. Find the important points that determine the area to be calculated.
First, we need to find the point where the graph crosses itself. It looks like this occurs at the point \((0,12)\) but we can't guarantee that just by looking at the graph. So we need to find \(t_1\) and \(t_2\) where the \(x\) and \(y\) values are equal.
At time \(t_1\), we have \( x_1 = {t_1}^3 - 3t_1 \) and \( y_1 = 4{t_1}^2\).
Similarly, at time \(t_2\), we have \( x_2 = {t_2}^3 - 3t_2 \) and \( y_2 = 4{t_2}^2\).
When the graph crosses, we have \( x_1 = x_2\) and \( y_1=y_2\). So we can write
\(\begin{array}{rcl} {t_1}^3 - 3t_1 & = & {t_2}^3 - 3t_2 \ 4{t_1}^2 & = & 4{t_2}^2 \end{array}\)
From the second equation, we may be tempted to write \( t_1 = t_2 \) only, but since \(t\) is squared, we also have the case where \( t_1 = -t_2 \). The time \( t_1 = t_2 \) does not help us since it is true by default. We need the case \( t_1 = -t_2 \). So we substitute \( t_1 = -t_2 \) into the first equation.
\( (-t_2)^3 - 3(-t_2) = {t_2}^3 - 3t_2 \)
For simplicity, we will drop the subscript temporarily in this equation and just use \(t\), i.e. \( t=t_2\).
\(\begin{array}{rcl} (-t)^3 - 3(-t) = t^3 - 3t \ -t^3 + 3t = t^3 - 3t \ 2t^3 - 6t & = & 0 \ 2t(t^2-3) & = & 0 \end{array}\)
Using the zero-product rule we have \( 2t = 0 \) and \(t^2-3=0 ~~ \to ~~ t^2 = 3 ~~ \to ~~ t = \pm \sqrt{3} \)
Now we will go back to the subscript to be clear. \( t_2 = \pm \sqrt{3} \)
So we have \(t_1 = -t_2 \) and \( t_2 = \pm \sqrt{3} \).
When \( t_2 = \sqrt{3}\), \( t_1 = -\sqrt{3} \) and when \( t_2 = -\sqrt{3}\), \( t_1 = \sqrt{3} \). So the graph crosses itself at \( t_1 = -\sqrt{3} \) and \(t_2 = \sqrt{3} \).
(Since both \(\sqrt{3}\) and \(-\sqrt{3}\) are covered here, it doesn't matter which set we choose. This may not always be the case, so be careful. We chose \( t_1 < t_2 \) just to be consistent with the idea that time is usually increasing.)
Let's find the \((x,y)\) point where the graph crosses itself. Let \( t = t_1 = -\sqrt{3}\) first.
\( x = {t_1}^3 - 3t_1 = (-\sqrt{3})^3 - 3(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0 ~~ \to ~~ x = 0 \)
\( y = 4{t_1}^2 = 4(-\sqrt{3})^2 = 4(3) = 12 ~~ \to ~~ y = 12 \)
So the point where the graph crosses itself is \((0,12)\). Just to confirm our work, let's find the point when \( t = t_2 = \sqrt{3} \).
\( x = (\sqrt{3})^3 - 3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0 \)
\( y = 4(\sqrt{3})^2 = 4(3) = 12 \)
So the t-values give us the same \( (x,y)=(0,12)\) point.

3. Determine how to set up the integrals in general to calculate the area.
Let's discuss how we are going to approach this. First, we may be tempted to find the area of half and then multiply by \(2\). The graph certainly looks symmetric about the y-axis but we would need to prove that it is and that's not easy to do. So we need to find both areas.

If we do the obvious and just integrate from \(t_1\) to \(t_2\), we will be calculating part of the area multiple times, since the integral calculates the area under the curve and above the x-axis. So we need to integrate from the left edge (the furthest x-value on the left) to the right edge to get the area shown in the left figure below. Then we need to subtract the areas beneath the curve (outside the region) shown in the right figure below. This will give us the area we are asked to calculate.

Our next step is to calculate the far left and far right points on the graph. To do that we notice that, at those points, the graph has vertical asymptotes, meaning that the slope is undefined. In the slope equation
\(\displaystyle{\frac{dy}{dx} = \frac{dy/dt}{dx/dt}}\)
\( dx/dt = 0 \) is where the slope is undefined (as long as \(dy/dt \neq 0 \)).
So we have \( x = t^3 - 3t ~~ \to ~~ dx/dt = 3t^2-3 \)
Setting \(dx/dt\) equal to zero and solving for \(t\), we find \( t = \pm 1 \).
We need to check that \(dy/dt \neq 0\) at these points.
\( y = 4t^2 ~~ \to ~~ dy/dt = 8t \) and at \( t=1 ~~ dy/dt = 8 \neq 0 \) and \( t=-1 ~~ dy/dt = -8 \neq 0 \). So, \( t = \pm 1 \) are the \(t\)-values when the graph has vertical asymptotes, which are the left and right boundaries of the area we are calculating.

The \((x,y)\) points are
\( t = -1: ~~ x = (-1)^3 - 3(-1) = -1+3 = 2 ~~ y = 4(-1)^2 = 4 ~~ (2,4) \)
\( t = 1: ~~ x = (1)^3 - 3(1) = 1-3 = -2 ~~ y = 4(1)^2 = 4 ~~ (-2,4) \)

Side Note: Isn't it interesting that as \(t\) sweeps from \( t = -\sqrt{3}\) to \( t = \sqrt{3}\), the graph is swept out clockwise? You need to know this to know how to correctly set up the integrals.

4. Set up the integrals and evaluate.
So now we have all the information we need to set up the integrals. Generally, the integral is
\(\displaystyle{\int_{t_1}^{t_2}{y(t)x'(t)dt}}\)

Area = \(\displaystyle{ \int_{-\sqrt{3}}^{-1}{4t^2(3t^2-3)dt} + \int_{1}^{\sqrt{3}}{4t^2(3t^2-3)dt} - \int_{1}^{-1}{4t^2(3t^2-3)dt} }\)

The first integral calculates the area under the top part of the curve on the right side of the y-axis down to the x-axis.
The second integral calculates the same area on the left side of the y-axis.
The last integral calculates the area below the bottom part of the curve on both sides of the y-axis and subtracts it from the other values, giving us the area enclosed by the curve. Notice that we need to sweep left to right for all the integrals. This is where the direction the curve is being swept is important.

All the integrals are the same expect for the limits of integration. So we will integrate once and then substitute the limits in. We may get some cancellation which will simplify our algebra. So we will watch for that.
\( \displaystyle{ \int{4t^2(3t^2-3)dt} = \int{12t^4-12t^2 dt} = 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right] + C }\)
We can drop the \(+C\) when we substitute the limits of integration since it will cancel in the subsequent calculations.

Area = \(\displaystyle{ 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right]_{-\sqrt{3}}^{-1} + 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right]_{1}^{\sqrt{3}} - 12 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right]_{1}^{-1} }\)

Area =\(\displaystyle{12 \left[ \frac{(-1)^5}{5} - \frac{(-1)^3}{3} \right] - 12 \left[ \frac{(-\sqrt{3})^5}{5} - \frac{(-\sqrt{3})^3}{3} \right]}\)
\(\displaystyle{ + 12 \left[ \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} \right] - 12 \left[ \frac{1}{5} - \frac{1}{3} \right] - 12 \left[ \frac{(-1)^5}{5} - \frac{(-1)^3}{3} \right] + 12 \left[ \frac{(1)^5}{5} - \frac{(1)^3}{3} \right] }\)
Let's factor out 12 and simplify the negative signs to see if we get any cancellation.
Area = \(\displaystyle{ 12 \left[ \frac{-1}{5} + \frac{1}{3} + \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} + \frac{(\sqrt{3})^5}{5} - \frac{(\sqrt{3})^3}{3} -\frac{1}{5} + \frac{1}{3} + \frac{1}{5} - \frac{1}{3} + \frac{1}{5} - \frac{1}{3} \right] }\)
So all the terms without roots cancel giving us
Area = \(\displaystyle{ 12 \left[ \frac{2}{5}(\sqrt{3})^5 - \frac{2}{3}(\sqrt{3})^3 \right] }\)
Simplifying and combining terms, we get Area = \(\displaystyle{ \frac{96}{5}\sqrt{3} }\)
Note:
Although we mentioned above that integrating from \(-\sqrt{3}\) to \( \sqrt{3}\) would not work, that is only partially true. In general cases, it would not work since integration is sensitive to direction, i.e. left to right gives you the positive area under the curve; right to left gives you the negative area under the curve. However, IN THIS PROBLEM ONLY, because of the direction the curve is swept, using one integral, sweeping from \(-\sqrt{3}\) to \( \sqrt{3}\) would give the correct answer. However, you would need to explain why. Do you see why?

Whew! That was a lot of work. I hope you didn't get lost in all the calculations. If you lost sight of the big picture, go back and scan the overview at the top of the solution to make sure you understand the main structure of this solution.

Final Answer

Area = \(\displaystyle{ \frac{96}{5}\sqrt{3} }\)

close solution

Log in to rate this practice problem and to see it's current rating.

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

Problem Statement

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

Final Answer

\(\displaystyle{6-\frac{15}{8}-2\ln(2)}\)

Problem Statement

Calculate the area enclosed by the line \(y=2.5\) and the parametric curve \(x=t-1/t\), \(y=t+1/t\).

Solution

In the video, he sets up the integral but does not evaluate it. Here are some of those steps.

\(\displaystyle{ \int_{1/2}^{2}{ \left[ 2.5-(t+1/2) \right]( 1+1/t^2)~dt } }\)

\(\displaystyle{ \int_{0.5}^{2}{ 2+2t^{-2} - t - t^{-1}~dt } }\)

\(\displaystyle{ \left[ 2t - \frac{2}{t} - \frac{t^2}{2} - \ln(t) \right]_{0.5}^{2} }\)

\(\displaystyle{ 6 - \frac{15}{8} - 2\ln(2) }\)

1376 video

video by PatrickJMT

Final Answer

\(\displaystyle{6-\frac{15}{8}-2\ln(2)}\)

close solution

Log in to rate this practice problem and to see it's current rating.

Calculate the area under one arc of the parametric curve \( x = r \theta - d \sin\theta \), \( y = r - d \cos\theta \).

Problem Statement

Calculate the area under one arc of the parametric curve \( x = r \theta - d \sin\theta \), \( y = r - d \cos\theta \).

Solution

1281 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

parametric calculus 17calculus youtube playlist

Here is a playlist of the videos on this page.

You CAN Ace Calculus

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

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

To bookmark this page and practice problems, log in to your account or set up a free account.

Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

calculus motivation - music and learning

Get great tutoring at an affordable price with Chegg. Subscribe today and get your 1st 30 minutes Free!

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

Practice

Join Amazon Student - FREE Two-Day Shipping for College Students

Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com!

Shop Amazon - Rent eTextbooks - Save up to 80%

Page Sections

How to Develop a Brilliant Memory Week by Week: 50 Proven Ways to Enhance Your Memory Skills

Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com!

Try AmazonFresh Free Trial

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.