5.7 Change of variables in multiple integrals (Page 4/13)

Calculus volume 3 Page 4 / 13

Considering the integral $\int_{0}^{1} \int_{0}^{\sqrt{1 - x^{2}}} (x^{2} + y^{2}) d y d x,$ use the change of variables $x = r cos θ$ and $y = r sin θ,$ and find the resulting integral.

$\int_{0}^{π / 2} \int_{0}^{1} r^{3} d r d θ$

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Notice in the next example that the region over which we are to integrate may suggest a suitable transformation for the integration. This is a common and important situation.

Changing variables

Consider the integral $\iint_{R} (x - y) d y d x,$ where $R$ is the parallelogram joining the points $(1, 2),$ $(3, 4), (4, 3),$ and $(6, 5)$ ( [link] ). Make appropriate changes of variables, and write the resulting integral.

A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3). — The region of integration for the given integral.

First, we need to understand the region over which we are to integrate. The sides of the parallelogram are $x - y + 1 = 0, x - y - 1 = 0,$ $x - 3 y + 5 = 0, and x - 3 y + 9 = 0$ ( [link] ). Another way to look at them is $x - y = −1, x - y = 1,$ $x - 3 y = −5,$ and $x - 3 y = 9 .$

Clearly the parallelogram is bounded by the lines $y = x + 1, y = x - 1, y = \frac{1}{3} (x + 5),$ and $y = \frac{1}{3} (x + 9) .$

Notice that if we were to make $u = x - y$ and $v = x - 3 y,$ then the limits on the integral would be $−1 \leq u \leq 1$ and $−9 \leq v \leq - 5 .$

To solve for $x$ and $y,$ we multiply the first equation by $3$ and subtract the second equation, $3 u - v = (3 x - 3 y) - (x - 3 y) = 2 x .$ Then we have $x = \frac{3 u - v}{2} .$ Moreover, if we simply subtract the second equation from the first, we get $u - v = (x - y) - (x - 3 y) = 2 y$ and $y = \frac{u - v}{2} .$

A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)/3, and y = (x + 5)/3. — A parallelogram in the $x y -plane$ that we want to transform by a change in variables.

Thus, we can choose the transformation

T (u, v) = (\frac{3 u - v}{2}, \frac{u - v}{2})

and compute the Jacobian $J (u, v) .$ We have

J (u, v) = \frac{\partial (x, y)}{\partial (u, v)} = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | = | \begin{array}{l} 3 / 2 & - 1 / 2 \\ 1 / 2 & - 1 / 2 \end{array} | = - \frac{3}{4} + \frac{1}{4} = - \frac{1}{2} .

Therefore, $| J (u, v) | = \frac{1}{2} .$ Also, the original integrand becomes

x - y = \frac{1}{2} [3 u - v - u + v] = \frac{1}{2} [3 u - u] = \frac{1}{2} [2 u] = u .

Therefore, by the use of the transformation $T,$ the integral changes to

\iint_{R} (x - y) d y d x = \int_{−9}^{−5} \int_{−1}^{1} J (u, v) u d u d v = \int_{−9}^{−5} \int_{−1}^{1} (\frac{1}{2}) u d u d v,

which is much simpler to compute.

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Make appropriate changes of variables in the integral $\iint_{R} \frac{4}{{(x - y)}^{2}} d y d x,$ where $R$ is the trapezoid bounded by the lines $x - y = 2, x - y = 4, x = 0, and y = 0 .$ Write the resulting integral.

$x = \frac{1}{2} (v + u)$ and $y = \frac{1}{2} (v - u)$ and $\int_{−4}^{4} \int_{−2}^{2} \frac{4}{u^{2}} (\frac{1}{2}) d u d v .$

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We are ready to give a problem-solving strategy for change of variables.

Problem-solving strategy: change of variables

Sketch the region given by the problem in the $x y -plane$ and then write the equations of the curves that form the boundary.
Depending on the region or the integrand, choose the transformations $x = g (u, v)$ and $y = h (u, v) .$
Determine the new limits of integration in the $u v -plane .$
Find the Jacobian $J (u, v) .$
In the integrand, replace the variables to obtain the new integrand.
Replace $d y d x$ or $d x d y,$ whichever occurs, by $J (u, v) d u d v .$

In the next example, we find a substitution that makes the integrand much simpler to compute.

Evaluating an integral

Using the change of variables $u = x - y$ and $v = x + y,$ evaluate the integral

\iint_{R} (x - y) e^{x^{2} - y^{2}} d A,

where $R$ is the region bounded by the lines $x + y = 1$ and $x + y = 3$ and the curves $x^{2} - y^{2} = −1$ and $x^{2} - y^{2} = 1$ (see the first region in [link] ).

As before, first find the region $R$ and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made ( [link] ).

On the left-hand side of this figure, there is a complex region R in the Cartesian x y-plane bounded by x squared minus y squared = negative 1, x squared minus y squared = 1, x + y = 3, and x + y = 1. Then there is an arrow from this graph to the right-hand side of the figure marked with x = (u + v)/2 and y = (v minus u)/2. On the right-hand side of this figure there is a simpler region S in the Cartesian u v-plane bounded by u v = negative 1, u v = 1, v = 1, and v = 3. — Transforming the region $R$ into the region $S$ to simplify the computation of an integral.

Given $u = x - y$ and $v = x + y,$ we have $x = \frac{u + v}{2}$ and $y = \frac{v - u}{2}$ and hence the transformation to use is $T (u, v) = (\frac{u + v}{2}, \frac{v - u}{2}) .$ The lines $x + y = 1$ and $x + y = 3$ become $v = 1$ and $v = 3,$ respectively. The curves $x^{2} - y^{2} = 1$ and $x^{2} - y^{2} = −1$ become $u v = 1$ and $u v = −1,$ respectively.

Thus we can describe the region $S$ (see the second region [link] ) as

S = {(u, v) | 1 \leq v \leq 3, \frac{−1}{v} \leq u \leq \frac{1}{v}} .

The Jacobian for this transformation is

J (u, v) = \frac{\partial (x, y)}{\partial (u, v)} = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | = | \begin{array}{l} 1 / 2 & - 1 / 2 \\ 1 / 2 & 1 / 2 \end{array} | = \frac{1}{2} .

Therefore, by using the transformation $T,$ the integral changes to

\iint_{R} (x - y) e^{x^{2} - y^{2}} d A = \frac{1}{2} \int_{1}^{3} \int_{−1 / v}^{1 / v} u e^{u v} d u d v .

Doing the evaluation, we have

\frac{1}{2} \int_{1}^{3} \int_{−1 / v}^{1 / v} u e^{u v} d u d v = \frac{4}{3 e} \approx 0.490 .

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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