Area integrals

In the previous section we defined line integrals over a curve, $\mathcal{C}$ . Here, we consider an integral of a function, $f(x, y)$ over a region $\mathcal{R}$ in the $x y$ -plane:

\mathcal{I}=\iint_{\mathcal{R}} f(x, y) d A

Area integrals are computed by converting them into repeated integrals. To do this, we need the region $\mathcal{R}$ to be enclosed from below and above by curves $y=p(x)$ and $y=q(x)$ , respectively, as follows:

\iint_{\mathcal{R}} f(x, y) d A=\int_{x=a}^{b}\left(\int_{p(x)}^{q(x)} f(x, y) d y\right) d x .

When carrying out the inner integral (i.e. with respect to $y$ ), the variable $x$ is treated as a constant. What we are really doing here, is dividing up the region $\mathcal{R}$ into small rectangles using a closely spaced mesh of vertical and horizontal lines and then taking the sum over all rectangles in the subdivision of $\mathcal{R}$ .

The order in which the terms are added (i.e. adding up rectangles in the vertical lines first and adding up the rectangles in the horizontal lines second or vice-versa) does not really matter provided that the function $f(x, y)$ is continuous in the region of interest (see Fubini's theorem). So, in Eq. (2.75), we integrate with respect to $y$ first from a lower bounding curve $y=p(x)$ to an upper bounding curve $y=q(x)$ and then integrate over $x$ from $x=a$ to $x=b$ . Similarly, we can have the region $\mathcal{R}$ be bounded by curves $x=a(y)$ and $x=b(y)$ on the left and right, respectively and then integrating with respect to $y$ from $y=p$ to $y=q$ . This is given by:

\iint_{\mathcal{R}} f(x, y) d A=\int_{y=p}^{q}\left(\int_{a(y)}^{b(y)} f(x, y) d x\right) d y .

Fubini's theorem

If $f(x, y)$ is continuous on the region given by the rectangle $\mathcal{R}=[a, b] \times[p, q]$ then,

\iint_{\mathcal{R}} f(x, y) d A=\int_{a}^{b}\left(\int_{p}^{q} f(x, y) d y\right) d x=\int_{p}^{q}\left(\int_{a}^{b} f(x, y) d x\right) d y .

We note the following properties: (a) $\iint_{\mathcal{R}}[f(x, y)+g(x, y)] d x d y=\iint_{\mathcal{R}} f(x, y) d x d y+\iint_{\mathcal{R}} g(x, y) d x d y$ ;

(b) $\iint_{\mathcal{R}} c f(x, y) d x d y=c \iint_{\mathcal{R}} f(x, y) d x d y$ , where $c$ is a constant;

(c) If the region $\mathcal{R}$ is complicated, the integral can usually be carried out by splitting the region into into two (or more) subregions $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ , then:

\iint_{\mathcal{R}} f(x, y) d x d y=\iint_{\mathcal{R}_{1}} f(x, y) d x d y+\iint_{\mathcal{R}_{2}} f(x, y) d x d y .

Example 2.10 Evaluate the following integral:

\mathcal{I}=\iint_{\mathcal{R}} \frac{x}{y} d x d y

where $\mathcal{R}$ is the shaded region given in Fig. 2.74,

Solution Figure 2.74 shows a semicircle of radius 2 (recall the equation of a circle of radius 2 is given in implicit form by $y^{2}+x^{2}=2^{2}$ ) and the shaded region, $\mathcal{R}$ is defined by:

\mathcal{R}: 1 \leq y \leq 2, \quad 0 \leq x \leq \sqrt{4-y^{2}} .

These inequalities may be substituted in Eq. (2.105) to give the following double integral:

\iint_{\mathcal{R}} \frac{x}{y} d x d y=\int_{y=1}^{2} \int_{x=0}^{\sqrt{4-y^{2}}} \frac{x}{y} d x d y .

We first evaluate the inner integral treating $y$ as a constant:

\begin{aligned} \int_{x=0}^{\sqrt{4-y^{2}}} \frac{x}{y} d x & =\left[\frac{x^{2}}{2 y}\right]_{0}^{\sqrt{4-y^{2}}} \\ & =\frac{4-y^{2}}{2 y} \\ & =\frac{2}{y}-\frac{y}{2} \end{aligned}

Next, we evaluate the outer integral which gives the result of the double integral:

\begin{aligned} \iint_{\mathcal{R}} \frac{x}{y} d x d y & =\int_{1}^{2}\left(\frac{2}{y}-\frac{y}{2}\right) d y \\ & =\left[2 \ln y-\frac{y^{2}}{4}\right]_{1}^{2}, \\ & =2 \ln 2-\frac{3}{4} . \end{aligned}

Figure 2.6: Integration over part of a semicircle (see Example 2.10).

Higher order integrals

Triple integrals can be calculated by the same limiting process by subdividing the volume $\mathcal{V}$ into a mesh of small boxes using planes parallel to the coordinate planes. Similarly to Eq. (2.75) [or Eq. (2.76)], we have:

\iiint_{\mathcal{V}} f(x, y, z) d x d y d z=\int_{x=a}^{b}\left[\int_{y=h_{1}(x)}^{h_{2}(x)}\left(\int_{z=g_{1}(x, y)}^{g_{2}(x, y)} f(x, y, z) d z\right) d y\right] d x .

Quadruple and higher integrals can be evaluated by similar repetitive techniques.

Change of variables

In single-variable calculus, sometimes we find that it is convenient to express an integral through a change of variable. For example, let us say we want to integrate $\int_{a}^{b} f(x) d x$ and we find that it is more convenient to do so by changing the $x$ 's into a different variable, say $u$ , through $x=g(u)$ . Then, by the substitution rule, we find that we can express the integral as:

\int_{a}^{b} f(x) d x=\int_{c}^{d} f(g(u)) \frac{d g}{d u} d u

note of course, that in going from $x$ 's to $u$ 's, we also need to change the limits of integration. In Eq. (2.81), we find $c$ and $d$ through $c=g^{-1}(a)$ and $d=g^{-1}(b)$ , respectively. We want to do something similar in the case of double integrals (and higher order integrals). Usually, the reason for changing variables is so that we can obtain an integral that we can carry out with the new variables. Another reason that is related to the area and surface integrals we looked at in the previous sections, is to convert the given region into a nicer region that we can work with. For example, if the given region is an ellipse, we might want to use a change of variable to transform the region into a disk (which is much nicer to deal with) or, if we have a scalene or an obtuse triangle, we might want to transform it into a right-angle triangle. Before we move on, let us do an example in which we use a transformation to change a region given in $x y$ -coordinates to $u v$ -coordinates.

Example 2.13 Apply the transformation $x=u / 2$ and $y=3 v$ on the ellipse $x^{2}+\frac{y^{2}}{36}=1$ to obtain a new region in $u v$ -coordinates.

Solution All we need to do here is plug the transformation into the equation for the ellipse:

\begin{aligned} \left(\frac{u}{2}\right)^{2}+\frac{1}{36}(3 v)^{2} & =1 \\ \frac{u^{2}}{4}+\frac{v^{2}}{4} & =1 \\ u^{2}+v^{2} & =4 . \end{aligned}

The new region is given by a disk of radius 2 .

In order to change variables in a double integral we need the Jacobian of the transformation. We use transformations $x=g(u, v)$ and $y=h(u, v)$ to go from the $x y$ -coordinate space to $u v$ -coordinate space. Let us relate back to Eq. (2.81); in using the substitution rule to change from $x$ - to $u$ -variable space in the single-variable case through the transformation $x=g(u)$ , we made use of the change of $x=g(u)$ with $u$ using $d g / d u$ . In the case of two variables we need to compute the following quantity:

\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right|

Note that Eq. (2.82) is the Jacobian determinant of the transformation $x=$ $g(u, v)$ and $y=h(u, v)$ . Definition 2.7 Suppose we want to integrate $f(x, y)$ over the region $\mathcal{R}$ . Under the transformation, $x=g(u, v)$ and $y=h(u, v)$ , the region becomes $\mathcal{S}$ and the integral becomes:

\iint_{\mathcal{R}} f(x, y) d x d y=\iint_{\mathcal{S}} f(x(u, v), y(u, v))|J| d u d v

where $J$ is the Jacobian determinant given by Eq. (2.82). Note that $|J|$ denotes the absolute value of $J$ . Example 2.14 Evaluate the following integral

\mathcal{I}=\iint_{\mathcal{R}} x+y d A

where $\mathcal{R}$ is the trapezoidal region with vertices $(0,0),(5,0),(5 / 2,5 / 2)$ and $(5 / 2,-5 / 2)$ , using the transformation $x=2 u+3 v$ and $y=2 u-3 v$ .

Solution Figure 2.7 shows the region $\mathcal{R}$ . Each one of the equations of the lines (the lines corresponding to different equations are shown in different colours) was found using 2 points connected by the line. Now let us use the transformation on each line. For $y=x$ , we have:

2 u-3 v=2 u+3 v, \rightarrow v=0 .

For $y=-x$ , we have:

2 u-3 v=-(2 u+3 v), \rightarrow u=0 .

For $y=-x+5$ , we have:

2 u-3 v=-(2 u+3 v)+5, \rightarrow u=5 / 4 .

For $y=x-5$ , we have:

2 u-3 v=(2 u+3 v)-5, \rightarrow v=5 / 6 .

The new region, $\mathcal{S}$ is given by a rectangle whose sides are given by the four equations $v=0, u=0, u=5 / 4$ and $v=5 / 6$ . It follows that $u$ and $v$ are defined within:

0 \leq u \leq \frac{5}{4}, \quad 0 \leq v \leq \frac{5}{6} .

We will be using Eq. (2.83) to carry out the double integral so the next step is to compute the Jacobian of the transformation:

J=\left|\begin{array}{cc} 2 & 3 \\ 2 & -3 \end{array}\right|=-12

Finally, the integral is given by:

\begin{aligned} \iint_{\mathcal{R}} d A & =\int_{0}^{5 / 6} \int_{0}^{5 / 4}[(2 u+3 v)+(2 u-3 v)]|-12| d u d v \\ & =\frac{125}{4} \end{aligned}

Figure 2.7: The trapezoidal region $\mathcal{R}$ in Example 2.14.

Consider now a situation in which instead of having a transformation involving $x=g(u, v)$ and $y=h(u, v)$ , we have

u=g(x, y), \quad v=h(x, y) .

In this case, we define $J$ , in an analogous manner to Eq. (2.82):

J=\frac{\partial(u, v)}{\partial(x, y)}=\left|\begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right|

Then, the transformation from $(u, v)$ to $(x, y)$ becomes

d u d v \rightarrow|J| d x d y

and the transformation from $(x, y)$ back $(u, v)$ is given by

d x d y \rightarrow\left|J^{\prime}\right| d u d v

hence,

\left|J^{\prime}\right|=\frac{1}{|J|}

Thus, we can now write down something analogous to Definition 2.7, expressed by Eq. (2.83) above:

\begin{aligned} \iint_{\mathcal{R}} f(x, y) d x d y & =\iint_{\mathcal{S}} f(x(u, v), y(u, v))\left|J^{\prime}\right| d u d v \\ & =\iint_{\mathcal{S}} f(x(u, v), y(u, v)) \frac{1}{|J|} d u d v \end{aligned}

As an example, consider the integral

\iint_{\mathcal{R}}\left(x^{2}+y^{2}\right) \exp \left(x^{2}-y^{2}\right) d x d y

If $u=x y$ and $v=x^{2}-y^{2}$ , then $J$ is given by

\begin{aligned} J & =\left|\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right|=\left|\begin{array}{cc} y & x \\ 2 x & -2 y \end{array}\right| \\ & =-2 y^{2}-2 x^{2} \end{aligned}

Thus, $|J|=2\left(x^{2}+y^{2}\right)$ from which it follows that

\left|J^{\prime}\right|=\frac{1}{|J|}=\frac{1}{2\left(x^{2}+y^{2}\right)} .

The integral can then be expressed as follows

\begin{aligned} \iint_{\mathcal{R}}\left(x^{2}+y^{2}\right) \exp \left(x^{2}-y^{2}\right) d x d y & =\iint_{\mathcal{S}}\left(x^{2}+y^{2}\right) \exp \left(x^{2}-y^{2}\right)\left|J^{\prime}\right| d u d v \\ & =\iint_{\mathcal{S}}\left(x^{2}+y^{2}\right) \exp \left(x^{2}-y^{2}\right) \frac{1}{|J|} d u d v \\ & =\iint_{\mathcal{S}}\left(x^{2}+y^{2}\right) \exp (v) \frac{1}{2\left(x^{2}+y^{2}\right)} d u d v \\ & =\iint_{\mathcal{S}} \frac{1}{2} \exp (v) d u d v, \end{aligned}

where $S$ is a region in $(u, v)$ space.

We can generalise these ideas to changing variables in higher integrals. For example, the $3 \times 3$ Jacobian determinant of the transformation $x=x(u, v, w)$ , $y=y(u, v, w)$ and $z=z(u, v, w)$ is given by:

J=\left|\begin{array}{lll} \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{array}\right|

By making use of Eq. (2.95), the variables in the triple integral are changed as follows:

\iiint_{\mathcal{V}} f(x, y, z) d x d y d z=\iiint_{\mathcal{W}} f(x(u, v, w), y(u, v, w), z(u, v, w))|J| d u d v d w

where $\mathcal{W}$ is the volume in the $u v w$ -space corresponding to the volume $\mathcal{V}$ in $x y z$ -space.

Conservative vector fields Surface integrals