# Multiple Integrals

• ### Introduction

The function $y=f(x)$ is a function of single variable $x$. $y$ is the dependent variable and $x$ is the independent variable. The indefinite integral

${\int ydx = g(x)}$

is a function, such that

${\frac {d}{dx} g(x) = y = f(x)}$

The definite integral

${\int \limits_a^b ydx}$

gives the area under the curve $y=f(x)$, bounded by the lines $x=a$, $x=b$ and the $X$ axis.

These are examples of single integrals, where only 1 independent variable is present. If the number of independent variables in a function is greater than 1, multiple integrals are used.

For functions with 2 independent variables, $f_1 (x,y)$, double integrals are evaluated and for functions with 3 independent variables, $f_2 (x,y,z)$, triple integrals are evaluated. It will soon become clear that

I) Single integrals are evaluated over a line.

II) Double integrals are evaluated over an area and

III) Triple integrals are evaluated over a 3D space.

• ### Double Integrals

Let $f(x,y)$ be a function of 2 independent variables $x$ and $y$, defined over a region $R$, bounded by a closed curve $C$. Let this region $R$ be divided into $n$ smaller regions, such that $r$th region has area $\delta A_r$. Let $P(x_r,y_r)$ be a point inside the $r$th region. Then the double integral over the region $R$ is defined as

${\iint_R f(x,y)dxdy = \lim \limits_{n \to \infty, \delta A \to 0} \sum \limits_{r=1}^{n} f(x_r,y_r) \delta A_r}$

A critical part in evaluating multiple integrals is choosing the upper and lower limits. For double integrals, the limits of the inner integral are to be expressed as functions of the other variable, which corresponds to the outer integral. For example, consider the figure below:

${\iint_R f(x,y)dxdy = \int \limits_{x=a}^{x=b} \int \limits_{y=f_1 (x)}^{y=f_2(x)} f(x,y)dydx}$

Thus, the inner integral is to be evaluated w.r.t. $y$, considering $x$ as a constant. Having obtained the inner integral, now, we should evaluate the outer integral as a single integral (the usual one). The limits of outer integral will be constants ($a$ and $b$ in this case).

If the limits of the inner integrals are functions of $y$, the outer integral will be w.r.t $y$ and the inner integral must be evaluated w.r.t. $x$.

Note that order of integration DOES NOT matter, as long as the limits are appropriately put.

I) Sometimes, it is required to change the order of integration, when the inner integral is impossible to integrate. In such cases, the limits are obtained by plotting the graph.

II) If the integrand involves terms of the form $x^2+y^2$ or $x^2y^2$, it becomes easier, if we transform the integrand to polar coordinates,

${x= r \ cos (\theta), y = r \ sin (\theta), r^2=x^2 + y^2, \theta = tan^{-1} \frac y x}$

Using Jacobians,

${dxdy = |J| d\theta dr = \begin {vmatrix} \frac {\partial (x,y)}{\partial (r, \theta)} \end {vmatrix} d \theta dr = r d \theta dr}$

Generally, the inner integral is obtained w.r.t. $r$.

III) If $f(x,y) =1$, we get the area of the region.

• ### Illustrated Examples

Question :

Evaluate the following over the region ${y \ge x^2, x \ge 1}$.

${\iint \frac {1}{x^4+y^2} dxdy}$

Solution :

The region of integration is plotted below :

Clearly, to the left of line ${x=1}$, the region is absent, hence the lower limit on ${x}$ will be ${1}$. The upper limit will be ${\infty}$. No restriction on the highest value of ${x}$.

Let’s now consider a strip parallel to ${Y}$ axis. It will first intersect the region at ${y=x^2}$. So, the lower limit on ${y}$ is ${x^2}$. The upper limit will be ${\infty}$ because no curve bounds the region in that direction. Hence, given integral will be

${\int \limits_{x=1}^{\infty} \int \limits_{y=x^2}^{\infty} \frac {1}{x^4+y^2} dydx = I \ say}$

Since inner limits are of ${y}$, let’s first integrate w.r.t. ${y}$.

${I = \int \limits_{x=1}^{\infty} \left [\frac {1}{x^2} tan^{-1} \left (\frac {y}{x^2} \right) \right]_{x^2}^{\infty}dx}$

${= \int \limits_{x=1}^{\infty} \frac {1}{x^2} \left[ \frac {\pi}{2} - \frac {\pi}{4} \right] dx}$

${= \frac {\pi}{4} \int \limits_1^{\infty} x^{-2}dx}$

${= \frac {\pi}{4} \left [\frac {x^{-2+1}}{-2+1} \right ]_{1}^{\infty}}$

${= \frac {\pi}{4} \times \frac {1}{-1} \times \left[\frac 1 x \right]_1^{\infty}}$

${= \frac {\pi}{4} \times -1 \times [0 - 1]}$

${= \frac {\pi}{4}}$

• ### Triple Integrals

As stated earlier, a triple integral is evaluated over a $3D \ space$. Let there be a function $f(x,y,z)$ defined over a closed region having volume $V$. Let the region be divided into $n$ subregions, such that the $r$th subregion has volume $\delta V_r$. Let $f(x_r,y_r,z_r$ be a point in $r$th subregion. Then the triple integral is given by

${\iiint \limits_{V} f (x,y,z) dxdydz = \lim \limits_{n \to \infty, \delta V_r \to 0} f(x_r,y_r,z_r) \delta V_r}$

The limits of integrals are equally important. The innermost integral (say w.r.t. $z$) has limits expressed in terms of functions of the other 2 variables, $x,y$. Once it is evaluated treating $x$ and $y$ as constants, it simplifies to a double integral problem and can be solved as explained earlier.

I) Transforming the integral into spherical polar coordinates requires the following replacement:

${dxdydz = |J|d \theta d \phi dr = r^2 sin (\theta) d \theta d \phi dr}$

II) Transforming the integral into cylindrical polar coordinates requires the following replacement:

${dxdydz = |J|d \rho d \phi dz = \rho d \rho d \phi dz}$

III) For a sphere $x^2 + y^2 + z^2 = a^2$, the limits are $r=0$ to $r = a$, $\theta = 0$ to $\theta = \pi$, $\phi = 0$ to $\phi = 2 \pi$

IV) For a hemisphere $x^2 + y^2 + z^2 = a^2, z \ge 0$, the limits are $r=0$ to $r = a$, $\theta = 0$ to $\theta = \frac \pi 2$, $\phi = 0$ to $\phi = 2 \pi$

IV) Dirichlet’s Theorem for triple integrals :

If $x+y+z \le 1$, then

${\iiint x^{a-1}y^{b-1} z^{c-1} dxdydz = \frac {\Gamma (a) \Gamma (b) \Gamma (c)}{\Gamma (a+b+c+1)}}$

• ### Areas and Volumes

I) When in a double integral, the function $f(x,y)=1$ for all values of $(x,y)$ in the region, the double integral gives the area of the region.

II) When in a triple integral, the function $f(x,y,z)=1$ for all $(x,y,z)$ in the closed region, the triple integral gives the volume of the region.

III) Other applications include obtaining center of mass, center of gravity, mean values, RMS values, moment of inertia etc.