# Parametric Integration

Many curves we integrate come in the form $$y = f (x)$$. For most curves, this is fine, but it is not always possible or convenient to write it like this. It is in this scenario where parametric coordinates are useful.

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## Recap of parametric coordinates

In this scenario, let us introduce a 'dummy' variable, usually denoted as t. We call this a dummy variable as it is an abstract concept that assigns a value to either an x or y coordinate, and isn't plotted.

This means that instead of having a function of the form $$y = f (x)$$, we represent a curve by $$y(t) = g(t)$$, $$x (t) = h (t)$$, where h and t are functions that describe the change of the x and y coordinates respectively.

A curve is described by $$y (t) = 2 (t)$$, $$x (t) = 2 (t)$$, $$0 < t < 2\pi$$.

Expressing the parametric curve as $$(x (t))^2 + (y (t))^2 = (2\cos{t})^2 + (2\sin{t})^2 = 4$$ $$\cos^2{t} + \sin^2{t} = 4$$, we see that it actually describes a circle of radius 4, or $$x^2 + y^2 = 4$$.

## Why does parametric integration work?

Normally, we expect to evaluate an integral of the form $$\int y (x) dx$$; however, we need to change this because our curve is not in the form $$y (x)$$. We use a modified version of the Chain Rule. We can replace dx with $$\frac{dx}{dt}dt$$ (you can think of this as the dt's cancelling. While this is not technically how they work, as $$\frac{dx}{dt}$$ is not strictly a fraction, we can treat it as one for operational purposes). This gives an integral of the form $$\int{y(t)\frac{dx(t)}{dt}dt}$$.

We must also remember to do with parametric integrals is switch limits. Suppose we have an integral of the form $$\int^b_a{f(x)dx}$$. We must also switch the limits, which results in the integral being given as$$\int^d_c{f(t)\frac{dx}{dt}dt}$$, where $$c = x^{-1}(a)$$ and $$d = x^{-1}(b)$$.

## Examples of parametric integration

At first glance, this can be a tricky topic to get your head around, so let's walk through a couple of examples to try and consolidate what we have said so far.

A curve is defined parametrically with $$x(t) = 2 -t$$ and $$y(t) = e^t - 1$$. Find the area enclosed by the x-axis, the line x = 0 and the curve.

First thing to do to work out where the curve crosses de x-axis and where the line x = 0 crosses the curve.

If the line crosses de x-axis, then the y-value will be zero. Solving this, we have $$e^t -1 = 0$$ which implies $$e^t = 1$$ and in turn t = 0. When x = 0, then $$2 - t = 0$$ which implies t = 2.

This means that we now have our limits, and we can start the integral. We have:

$\int^0_2{(e^t -1)} \cdot \frac{d}{dt}(2 - t) \cdot dt = - \int^0_2{(e^t - 1)} dt = \int^0_2{(e^t - 1) dt}$

where we switch the limits to change the sign.

This then equals $$[e^t - t]_{t=0}^{t=2} = [(e^2 - 2) - (1-0)] = e^2 - 3$$.

Using parametric integration, find the area of the circle defined as $$x(t) = -3\cos(t), y(t) = 3\sin(t), 0 < t < 2\pi$$.

By the formula for parametric integration, we have:

$\int^{2\pi}_0 {3\sin(t) \cdot \frac {d}{dt} (-3 \cos (t))dt} = 9 \int^{2\pi}_0 \sin^2(t)dt$.

We now need to use a double angle formula here, and we can use the result $$2(t) = \frac{1}{2}(1 - \cos(2t))$$.

Filling this in, we get $$\frac{9}{2} \int^{2\pi}_0{(1-\cos(2t))dt} = \frac{9}{2}[t - \frac{1}{2} \sin(2t)]^{t = 2\pi}_{t = 0} = 9\pi$$, which is what we would expect of a circle with a radius of 3.

Exam style question

Suppose we have a curve that has been defined parametrically, with $$x(t) = 3\cos(4t)$$ and $$y(t) = 6 \sin(8t)$$, with $$0 < t < \frac{\pi}{8}$$.

i) Find any turning points of the curve.

ii) Find the area under the curve.

i) For a turning point, then $$\frac{dy}{dx}$$ must be equal. By the chain rule,

$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt}(\frac{dx}{dt})^{-1}$

We can now use the standard formulae for derivatives of trigonometric functions to find these results.

$$\frac{dy}{dt} = 48 \cos(8t)$$ and $$\frac{dx}{dt} = -12 \sin(4t)$$, which in turn will give $$\frac{dy}{dx} = \frac{48 \cos(8t)}{-12 \sin(4t)}. We can then solve this equal to zero to find the t value of the turning point. For this to equal zero, then the numerator must equal zero, which implies that \(\cos(8t) = 0$$.

This means that $$8t = \frac{\pi}{2} + n\pi, n \epsilon N$$, which we further reduce to $$t = \frac{\pi}{16} + \frac{n\pi}{8}, n \epsilon N$$.

The only value of t here which satisfies the $$0 < t < 8\pi$$ is $$t = \frac{\pi}{16}$$.

This gives the x coordinate of the turning point as $$3\cos(\frac{\pi}{4}) = \frac{3\sqrt{2}}{2}$$, and the y coordinate as $$6\sin(\frac{\pi}{2}) = 6$$.

First, let's work out. the direction of our limits. $$x(0) = 3$$ and $$x(\frac{\pi}{8}) = 0$$, which means the area under the curve is given by

$\int^0_{\frac{\pi}{8}}{y(t) \cdot \frac{dx}{dt}\cdot dt} = \int^0_{\frac{\pi}{8}}{6 \sin(8t)(-12\sin(4t))dt} = 72\int^{\frac{\pi}{8}}_0{\sin(8t) \sin(4t)dt}$

where we "flip the limits" to get rid of the negative sign. We can use a double angle formula to help us solve this integral.

We know $$\sin(2x) = 2\sin(x)\cos(x)$$. This implies that $$\sin(8x) = 2 \cdot \sin(4x) \cos(4x)$$.

Filling this in, we get the integral $$144 \int^{\frac{\pi}{8}}_0 \sin^2(4t)\cos(4t)dt$$.

Since $$\int \sin(t) = \cos(t)$$, it seems intuitive that this is best suited to an integration by substitution.

Let us take $$u = \sin(4t)$$ which implies $$\frac{du}{dt} = 4\cos(4t)$$, so $$dt = \frac{du}{4\cos(4t)}$$.

As this is a definite integral, we also need to change the limits.

$$u_1 = \sin(4 \cdot 0) = 0$$ and $$u_2 = \sin(4 \cdot \frac{\pi}{8}) = 1$$.

Filling this in, we can say that $$144 \int_0^{\frac{\pi}{8}} \sin^2(4t) \cos(4t) dt = 36 \int^1_0 u^2 du$$.

This is a straightforward integral and can be evaluated directly.

$$36 \int^1_0 u^2 du = 12[u^3]^{u = 1}_{u = 0} = 12[1-0] = 12$$.

## Parametric Integration - Key takeaways

• The formula for parametric integration is given as $$\int{y(t)\frac{dx(t)}{dt}dt}$$.

• We must remember to switch limits when changing from x coordinates to t coordinates.

#### Flashcards in Parametric Integration 12

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How do you integrate parametric equations?

Integrate parametrically using the formula  ∫y(t).dx(t/)dt.dt

How do you find the limits of integration for parametric equations?

Use the inverse formula for x(t) to find the t limits from the x limits.

Why does parametric integration work?

Parametric integration works as we introduce a dummy variable which essentially shifts our curve to a different plane where it is easier to integrate.

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