Active Calculus - Multivariable

Subsection 12.8.1 Introduction

   

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Preview Activity 12.8.1.

(a)

Consider the double integral

\begin{equation} I = \iint_D x^2+y^2 \, dA \tag{12.8.1} \end{equation}

where \(D\) is the upper half of the unit disk (\(y \geq 0\)). Remember the polar conversion equations:

\begin{equation} x = r \cos(\theta) \quad \text{ and } \quad y = r \sin(\theta)\tag{12.8.2} \end{equation}

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(a)

Write the double integral \(I\) given in Equation (12.8.1) as an iterated integral in rectangular coordinates.

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(b)

Write the double integral \(I\) given in Equation  (12.8.1) as an iterated integral in polar coordinates.

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(c)

Draw a plot of the region of integration \(D\) in the \(xy\)-plane and label the four “corners” of the polar coordinates (if \(D\) is described by \(a \leq r \leq b , c \leq \theta \leq d\text{,}\) then the corners will be \((r,\theta)=(a,c),(a,d),(b,c),(b,d)\)).

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Hint.

Two of your “corners” will be at the same point

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(b)

Suppose you are calibrating a new radar system for your self driving car. This radar system is mounted at the center of your front bumper and can detect objects up to 45 degrees to either side of the forward direction at distances from 1 meter to 3 meters.

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(a)

On Figure 12.8.1, draw a plot of \(S\text{,}\) the region that your radar system can detect objects. Assume your car’s front bumper is centered at the origin of your coordinate system and the car is facing the positive \(y\)-direction. You should label the four corners of \(S\) in terms of both rectangular and polar coordinates.

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(b)

On Figure 12.8.2, plot the rectangle on the \(r\theta\)-plane that corresponds to the bounds \(a \leq r \leq b \quad c \leq \theta \leq d\) for your radar system. Label the four corners of your figure with their \(r\) and \(\theta\) coordinates.

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Subsection 12.8.2 Change of Variables in Polar Coordinates

Example 12.8.3.

The vertices of the polar rectangle \(P\) given by \([r_1, r_2] \times [\theta_1, \theta_2]\) are transformed into the vertices of a closed and bounded region \(P'\) in rectangular coordinates. If we view the standard coordinate system as having the horizontal axis represent \(r\) and the vertical axis represent \(\theta\text{,}\) then the polar rectangle \(P\) appears to us at left in Figure 12.8.4. The image \(P'\) of the polar rectangle \(P\) under the transformation given by (12.8.2) is shown at right in Figure 12.8.4. We thus see that there is a correspondence between a simple region (a traditional, right-angled rectangle) and a more complicated region (a fraction of an annulus) under the function \(T\) given by \(T(r, \theta) = (r\cos(\theta), r\sin(\theta))\text{.}\)

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Subsection 12.8.3 General Change of Coordinates

   

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Activity 12.8.2.

In this activity, we will be looking at what happens to the region \(D\text{,}\) as shown in Figure 12.8.5, under the change of coordinates given by

\begin{equation*} x=\sqrt{st} \quad \text{ and } \quad y =\sqrt{\frac{s}{t}} \end{equation*}

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(a)

Find the coordinates of the four corner points of the region \(D\text{.}\)

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(b)

Given the form of the change of variable equations above, it is not obvious how to change \((x,y)\) points into \((s,t)\) points. Show that the following equations are satisfied by our change of variable equations:

\begin{equation*} yx=s \quad \text{and} \quad \frac{y}{x}=\frac{1}{t} \end{equation*}

Use these new relationships between our coordinates to convert each of these four corner points from \(xy\)-coordinates to \(st\)-coordinates and draw these points on Figure 12.8.6.

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(c)

We hope that the curved boundaries of the region \(D\) get transformed to horizontal and vertical lines in the \(st\)-plane, as is suggested by the result of the previous task. Let’s look at how each of the boundary curves gets transformed by our change of variables formula.

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Use the change of variables equations to write \(y=2x\) in terms of \(s\) and \(t\) coordinates and solve for either \(s\) or \(t\) (as would make sense). Plot the resulting line on Figure 12.8.6.

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(d)

Write each of the other boundary curves of \(D\) (\(y=\frac{2}{x}, y=\frac{1}{2x}, y=\frac{x}{2}\)) in terms of \(s\) and \(t\) coordinates, solve for either \(s\) or \(t\) (as would make sense), and plot the resulting lines on Figure 12.8.6.

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(e)

We have now confirmed that our irregular region \(D\) gets mapped to a rectangle in the \(st\)-plane by the change of variables formula given. Some areas of the region \(D\) needed to be stretched more than others in order to straighten out our sides into the nice rectangle you have drawn on Figure 12.8.6. Write a few sentences to describe what part of \(D\) you think is stretched the most in the transformation from \(xy\)-coordinates to \(st\)-coordinates.

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Example 12.8.7.

If we look at the integral

\begin{equation*} \int_0^2 2x(x^2+1)^3\, dx, \end{equation*}

and consider the change of variable \(u = x^2+1\) with \(du = 2x \, dx\text{,}\) then \(x = 0\) implies \(u = 1\) and \(x = 2\) implies \(u = 5\text{.}\) So we have transformed the original integral in \(x\) into a new integral in \(u\text{.}\) In particular,

\begin{equation*} \int_0^2 (x^2+1)^3 (2x\, dx) = \int_1^5 u^3 \, du \end{equation*}

which we can visualize with the following plots of area under curves.

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Both of these shaded regions have the same area but you can see that the scales on the vertical and horizontal axes changed. The vertical axis on each graph corresponds to the output of the function being integrated and the transformation comes from transforming the integrand. The bounds on the integral come from the conversion of the interval \([0,2]\) in \(x\)-coordinates to the interval \([1,5]\) in \(u\)-coordinates. The final part of this substitution/change of variable problem is converting the differential.

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Proof.

We partition a rectangular domain in the \(st\) system into subrectangles. Let \(T = [a, b] \times [a+\Delta s, b+\Delta t]\) be one of these subrectangles. Then we transform this into a region \(T'\) in the standard \(xy\) Cartesian coordinate system. The region \(T'\) is called the image of \(T\text{;}\) the region \(T\) is the pre-image of \(T'\text{.}\) Although the sides of this \(xy\) region \(T'\) aren’t necessarily straight (linear), we will approximate the element of area \(dA\) for this region with the area of the parallelogram whose sides are given by the vectors \(\vv\) and \(\vw\text{,}\) where \(\vv\) is the vector from \((x(a, b), y(a, b))\) to \((x(a + \Delta s, b), y(a + \Delta s, b))\text{,}\) and \(\vw\) is the vector from \((x(a, b), y(a, b))\) to \((x(a, b + \Delta t), y(a, b + \Delta t))\text{.}\)

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An example of an image \(T'\) in the \(xy\)-plane that results from a transformation of a rectangle \(T\) in the \(st\)-plane is shown in Figure 12.8.11.

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The components of the vector \(\vv\) are

\begin{align*} \vv \amp = \left\langle x(a+ \Delta s, b) - x(a,b), y(a+ \Delta s, b) - y(a,b), 0 \right\rangle \end{align*}

and for \(\vw\) the components are

\begin{align*} \vw \amp = \left\langle x(a, b+ \Delta t) - x(a,b), y(a, b+ \Delta s) - y(a,b), 0 \right\rangle. \end{align*}

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Slightly rewriting \(\vv\) and \(\vw\text{,}\) we have

\begin{align*} \vv \amp = \left\langle \frac{x(a+ \Delta s, b) - x(a,b)}{\Delta s}, \frac{y(a+ \Delta s, b) - y(a,b)}{\Delta s}, 0 \right\rangle \Delta s, \ \mbox{and}\\ \vw \amp = \left\langle \frac{x(a, b+ \Delta t) - x(a,b)}{\Delta t}, \frac{y(a, b+ \Delta s) - y(a,b)}{\Delta t}, 0 \right\rangle \Delta t. \end{align*}

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For small \(\Delta s\) and \(\Delta t\text{,}\) the definition of the partial derivative tells us that

\begin{equation*} \vv \approx \left\langle \frac{\partial x}{\partial s}(a,b), \frac{\partial y}{\partial s}(a,b), 0 \right\rangle \Delta s \ \ \ \ \ \text{ and } \ \ \ \ \ \vw \approx \left\langle \frac{\partial x}{\partial t}(a,b), \frac{\partial y}{\partial t}(a,b), 0 \right\rangle \Delta t. \end{equation*}

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Recall that the area of the parallelogram with sides \(\vv\) and \(\vw\) is the length of the cross product of the two vectors, \(|\vv \times \vw|\text{.}\) From this, we observe that

\begin{align*} \vv \times \vw \amp \approx \left\langle \frac{\partial x}{\partial s}(a,b), \frac{\partial y}{\partial s}(a,b), 0 \right\rangle \Delta s \times \left\langle \frac{\partial x}{\partial t}(a,b), \frac{\partial y}{\partial t}(a,b), 0 \right\rangle \Delta t\\ \amp = \left\langle 0, \ 0, \ \frac{\partial x}{\partial s}(a,b) \frac{\partial y}{\partial t}(a,b) - \frac{\partial x}{\partial t}(a,b) \frac{\partial y}{\partial s}(a,b) \right\rangle \Delta s \, \Delta t. \end{align*}

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Finally, by computing the magnitude of the cross product, we see that

\begin{align*} |\vv \times \vw| \amp \approx \left|\left\langle 0,0, \frac{\partial x}{\partial s}(a,b) \frac{\partial y}{\partial t}(a,b) - \frac{\partial x}{\partial t}(a,b) \frac{\partial y}{\partial s}(a,b) \right\rangle \Delta s \, \Delta t\right|\\ \amp = \left|\frac{\partial x}{\partial s}(a,b) \frac{\partial y}{\partial t}(a,b) - \frac{\partial x}{\partial t}(a,b) \frac{\partial y}{\partial s}(a,b)\right| \Delta s \, \Delta t. \end{align*}

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Therefore, as the number of subdivisions increases without bound in each direction, \(\Delta s\) and \(\Delta t\) both go to zero, and we have

\begin{equation} dA = \left|\frac{\partial x}{\partial s} \frac{\partial y}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial y}{\partial s} \right| ds \, dt.\tag{12.8.3} \end{equation}

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Subsection 12.8.4 Change of Variables in a Triple Integral

Subsection 12.8.5 Summary

• If an integral is described in terms of one set of variables, we may write that set of variables in terms of another set of the same number of variables. If the new variables are chosen appropriately, the transformed integral may be easier to evaluate.

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• The Jacobian is a scalar function that relates the area or volume element in one coordinate system to the corresponding element in a new system determined by a change of variables.

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