## 17Calculus Parametric Equations - Area

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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

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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

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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 $$x = t^3 - 3t$$, $$y = 4t^2$$.

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.

Area = $$\displaystyle{ \frac{96}{5}\sqrt{3} }$$

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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 $$x=t-1/t$$, $$y=t+1/t$$.

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

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

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

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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

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