5.7 Change of variables in multiple integrals (Page 6/13)

Calculus volume 3 Page 6 / 13

Let’s try another example with a different substitution.

Evaluating a triple integral with a change of variables

Evaluate the triple integral

\int_{0}^{3} \int_{0}^{4} \int_{y / 2}^{(y / 2) + 1} (x + \frac{z}{3}) d x d y d z

in $x y z -space$ by using the transformation

u = (2 x - y) / 2, v = y / 2, and w = z / 3 .

Then integrate over an appropriate region in $u v w -space .$

As before, some kind of sketch of the region $G$ in $x y z -space$ over which we have to perform the integration can help identify the region $D$ in $u v w -space$ ( [link] ). Clearly $G$ in $x y z -space$ is bounded by the planes $x = y / 2, x = (y / 2) + 1, y = 0,$ $y = 4,$ $z = 0, and z = 4 .$ We also know that we have to use $u = (2 x - y) / 2, v = y / 2, and w = z / 3$ for the transformations. We need to solve for $x, y, and z .$ Here we find that $x = u + v,$ $y = 2 v,$ and $z = 3 w .$

Using elementary algebra, we can find the corresponding surfaces for the region $G$ and the limits of integration in $u v w -space .$ It is convenient to list these equations in a table.

Equations in $x y z$ for the region $D$	Corresponding equations in $u v w$ for the region $G$	Limits for the integration in $u v w$
$x = y / 2$	$u + v = 2 v / 2 = v$	$u = 0$
$x = y / 2$	$u + v = (2 v / 2) + 1 = v + 1$	$u = 1$
$y = 0$	$2 v = 0$	$v = 0$
$y = 4$	$2 v = 4$	$v = 2$
$z = 0$	$3 w = 0$	$w = 0$
$z = 3$	$3 w = 3$	$w = 1$

On the left-hand side of this figure, there is a box G with sides 1, 2, and 1 along the u, v, and w axes, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u + v, y = 2v, and z = 3w. On the right-hand side of this figure there is a region D in xyz space that is a rotated box with sides 1, 4, and 3 along the x, y, and z axes. The rear plane is marked x = y/2 or y = 2x. The front plane is marked x = y/2 + 1 or y = 2x minus 2. — The region $G$ in $u v w -space$ is transformed to region $D$ in $x y z -space .$

Now we can calculate the Jacobian for the transformation:

J (u, v, w) = | \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} | = | \begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix} | = 6 .

The function to be integrated becomes

f (x, y, z) = x + \frac{z}{3} = u + v + \frac{3 w}{3} = u + v + w .

We are now ready to put everything together and complete the problem.

\begin{matrix} \int_{0}^{3} \int_{0}^{4} \int_{y / 2}^{(y / 2) + 1} (x + \frac{z}{3}) d x d y d z \\ = \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} (u + v + w) | J (u, v, w) | d u d v d w = \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} (u + v + w) | 6 | d u d v d w \\ = 6 \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} (u + v + w) d u d v d w = 6 {\int_{0}^{1} \int_{0}^{2} [\frac{u^{2}}{2} + v u + w u]}_{0}^{1} d v d w \\ = 6 \int_{0}^{1} \int_{0}^{2} (\frac{1}{2} + v + w) d v d w = 6 {\int_{0}^{1} [\frac{1}{2} v + \frac{v^{2}}{2} + w v]}_{0}^{2} d w \\ = 6 \int_{0}^{1} (3 + 2 w) d w = 6 {[3 w + w^{2}]}_{0}^{1} = 24. \end{matrix}

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Let $D$ be the region in $x y z -space$ defined by $1 \leq x \leq 2, 0 \leq x y \leq 2, and 0 \leq z \leq 1 .$

Evaluate $\underset{D}{∭} (x^{2} y + 3 x y z) d x d y d z$ by using the transformation $u = x, v = x y,$ and $w = 3 z .$

$\int_{0}^{3} \int_{0}^{2} \int_{1}^{2} (\frac{v}{3} + \frac{v w}{3 u}) d u d v d w = 2 + ln 8$

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Key concepts

A transformation $T$ is a function that transforms a region $G$ in one plane (space) into a region $R$ in another plane (space) by a change of variables.
A transformation $T : G \to R$ defined as $T (u, v) = (x, y)$ $(or T (u, v, w) = (x, y, z))$ is said to be a one-to-one transformation if no two points map to the same image point.
If $f$ is continuous on $R,$ then $\iint_{R} f (x, y) d A = \iint_{S} f (g (u, v), h (u, v)) | \frac{\partial (x, y)}{\partial (u, v)} | d u d v .$
If $F$ is continuous on $R,$ then

$\begin{matrix} \underset{R}{∭} F (x, y, z) d V & = \underset{G}{∭} F (g (u, v, w), h (u, v, w), k (u, v, w)) | \frac{\partial (x, y, z)}{\partial (u, v, w)} | d u d v d w \\ = \underset{G}{∭} H (u, v, w) | J (u, v, w) | d u d v d w . \end{matrix}$

In the following exercises, the function $T : S \to R, T (u, v) = (x, y)$ on the region $S = {(u, v) | 0 \leq u \leq 1, 0 \leq v \leq 1}$ bounded by the unit square is given, where $R \subset R^{2}$ is the image of $S$ under $T .$

Justify that the function $T$ is a $C^{1}$ transformation.
Find the images of the vertices of the unit square $S$ through the function $T .$
Determine the image $R$ of the unit square $S$ and graph it.

$x = 2 u, y = 3 v$

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$x = \frac{u}{2}, y = \frac{v}{3}$

a. $T (u, v) = (g (u, v), h (u, v)), x = g (u, v) = \frac{u}{2}$ and $y = h (u, v) = \frac{v}{3} .$ The functions $g$ and $h$ are continuous and differentiable, and the partial derivatives $g_{u} (u, v) = \frac{1}{2},$ $g_{v} (u, v) = 0, h_{u} (u, v) = 0 and h_{v} (u, v) = \frac{1}{3}$ are continuous on $S;$ b. $T (0, 0) = (0, 0),$ $T (1, 0) = (\frac{1}{2}, 0), T (0, 1) = (0, \frac{1}{3}),$ and $T (1, 1) = (\frac{1}{2}, \frac{1}{3});$ c. $R$ is the rectangle of vertices $(0, 0), (\frac{1}{2}, 0), (\frac{1}{2}, \frac{1}{3}), and (0, \frac{1}{3})$ in the $x y -plane;$ the following figure.
A rectangle with one corner at the origin, horizontal length 0.5, and vertical height 0.34.

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$x = u - v, y = u + v$

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$x = 2 u - v, y = u + 2 v$

a. $T (u, v) = (g (u, v), h (u, v)), x = g (u, v) = 2 u - v,$ and $y = h (u, v) = u + 2 v .$ The functions $g$ and $h$ are continuous and differentiable, and the partial derivatives $g_{u} (u, v) = 2,$ $g_{v} (u, v) = −1,$ $h_{u} (u, v) = 1,$ and $h_{v} (u, v) = 2$ are continuous on $S;$ b. $T (0, 0) = (0, 0),$ $T (1, 0) = (2, 1),$ $T (0, 1) = (−1, 2),$ and $T (1, 1) = (1, 3);$ c. $R$ is the parallelogram of vertices $(0, 0), (2, 1), (1, 3), and (−1, 2)$ in the $x y -plane;$ see the following figure.
A square of side length square root of 5 with one corner at the origin and another at (2, 1).

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Questions & Answers

calculate molarity of NaOH solution when 25.0ml of NaOH titrated with 27.2ml of 0.2m H2SO4

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Read Chapter 6, section 5

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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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