Double Integration Engineering Maths, Btech first year

Double Integration

Before we jump into Double Integration, if you need to revise some basic concepts of Integration Click Here.

In normal Integration we would perform integration w.r.t. only one variable; in Double Integration we will perform integration w.r.t. two variables.

Here we are going to study three types of Double Integration:
Type 1 : With constant limits
Type 2 : With variable limits
Type 3 : No limits $⟶$ we would have to find out from the question

Type-1 : With constant limits :

$${\int }_0^4{\int }_0^{10}xydxdy$$ Here the confusion is- which variable has which limits, so below is the rule: $$\text{Inner Variable}⟶\text{Inner Limit}$$ $$\text{Outer Variable}⟶\text{Outer Limit}$$ In this problem $x$ has limits 0 to 10, $y$ has limits 0 to 4.
And there is no rule for which variable has to be integrated first so you can either first integrate $x$ and then $y$ or vice-versa. Let's integrate the above problem in both orders:

1) First $x$, then $y$

$${\int }_0^4{\left[\frac{x^2}{2}\right]}_0^{10}ydy=50{\int }_0^{4}ydy$$ $$=50{\left[\frac{y^2}{2}\right]}_0^4=50\times 8=400$$

2) First $y$, then $x$ :

$${\int }_0^{10}x{\left[\frac{y^2}{2}\right]}_0^{4}dx=8{\int }_0^{10}xdx$$ $$=8{\left[\frac{x^2}{2}\right]}_0^4=8\times 50=400$$ As you can see, answers in both the cases are same.

Type-2 : With variable limits

$${\int }_0^1{\int }_{x^2}^{x}xydxdy$$ Here the rule that you used in first type is not applicable. In this type, one limit consists of $x$ variable, so the question is: to which variable does it belong? The answer is simple, consider this: does $x=x$ make any sense to you? No. Does $y=x$ make sense to you? Yes. Why? Because in first equation $x$ was written in terms of $x$ which just doesn't make any sense and in the second equation, $x$ was written in terms of $y$ which is obvious. So from this you can conclude that if the limit consists of $x$ variable, then it must belong to $y$ and if the limit consists of $y$ variable, then it must belong to $x$.
Also here the order of integration matters: the one with variable limits will be integrated first. So let's solve the above problem:

$${\int }_0^1{\left[\frac{y^2}{2}\right]}_{x^2}^{x}xdx$$ $$={\int }_0^1[\frac{x^2}{2}-\frac{x^4}{2}]xdx$$ $$=\frac{1}{2}{\int }_0^1(x^3-x^5)dx$$ $$=\frac{1}{2}{[\frac{x^4}{4}-\frac{x^6}{6}]}_0^1$$ $$=\frac{1}{2}(\frac{1}{4}-\frac{1}{6})=\frac{1}{24}$$

Type 3 : No limit

In this type of problems, we won't be given limits, instead we will be given equations of lines or curves which we will use to determine the region in cartesian plane over which we will integrate the given integral. Let's see an example :

Sample Problem 1

Evaluate $\iint xydxdy$ over positive quadrant of the circle $x^2+y^2=a^2$

Solution

In these kind of problems, we follow the following steps:
1) Trace the curve using the equations of curves given in the question
2) Identify the region over which we will integrate
3) Find the limits
4) Integrate
Step-1 : Trace the curve $$x^2+y^2=a^2$$ It is a circle of radius $a$ units and center (0,0).

Step-2 : identify the region
Here it is given the positive quadrant so,

Step-3 : Find the limits
Now comes the main part. There are two ways to find limits, Horizontal Strip or Vertical Strip. Now consider this, integration is all about finding area, so how do you find area of such region? Simple, take a strip, and drag it to cover the whole region. Here you have two ways: Horizontal Strip or Vertical Strip.

If you take a horizontal strip, it will be parallel to the x-axis and vertical strip will be parallel to y-axis. The both ends of the strip will be the limits i.e. the equation of the curves they are touching: if horizontal strip, then limits will belong to $x$, if vertical strip, then the limits will belong to $y$. So let's solve it both the ways:

Horizontal Strip : For limits of $x$:
One end is on the y-axis so the equation is $x=0$
The other end lies on the circle so the equation becomes: $$x^2+y^2=a^2$$ $$x^2=a^2-y^2$$ $$x=\pm \sqrt{a^2-y^2}$$ Since it is the positive quadrant so $$\boxed{{\displaystyle x=\sqrt{a^2-y^2}}}$$ One thing to consider while writing limits in integral: in horizontal strip, we go from left to right and in vertical strip, from bottom to top
Step-4 : Now evaluating integral $${\int }_0^a{\int }_0^{\sqrt{a^2-y^2}}xydxdy$$ $$={\int }_0^a{\left[\frac{x^2}{2}\right]}_0^{\sqrt{a^2-y^2}}ydy$$ $$={\int }_0^a\left[\frac{a^2-y^2}{2}\right]ydy$$ $$=\frac{1}{2}{\int }_0^a(a^{2}y-y^3)dy$$ $$=\frac{1}{2}{[\frac{a^2{y}^2}{2}-\frac{y^4}{4}]}_0^a$$ $$=\frac{1}{2}[\frac{a^4}{2}-\frac{a^4}{4}]$$ $$=\boxed{{\displaystyle \frac{a^4}{8}}}$$ Vertical Strip : For limits of $y$ :
One end is on the x-axis so the equation is $y=0$
The other end lies on the circle so the equation becomes: $$x^2+y^2=a^2$$ $$y^2=a^2-x^2$$ $$y=\pm \sqrt{a^2-x^2}$$ Since it is the positive quadrant so $$\boxed{{\displaystyle y=\sqrt{a^2-x^2}}}$$ Now evaluating integral $${\int }_0^a{\int }_0^{\sqrt{a^2-x^2}}xydxdy$$ $$={\int }_0^a{\left[\frac{y^2}{2}\right]}_0^{\sqrt{a^2-x^2}}xdx$$ $$={\int }_0^a\left[\frac{a^2-x^2}{2}\right]xdx$$ $$=\frac{1}{2}{\int }_0^a(a^{2}x-x^3)dx$$ $$=\frac{1}{2}{[\frac{a^2{x}^2}{2}-\frac{x^4}{4}]}_0^a$$ $$=\frac{1}{2}[\frac{a^4}{2}-\frac{a^4}{4}]$$ $$=\boxed{{\displaystyle \frac{a^4}{8}}}$$

Sample Problem 2

Evaluate $\iint xy(x+y)dxdy$ over the area bounded between the curves $y=x$ and $y=x^2$.

Solution

First trace the curve, second identify the region.
$$y=x^2$$ $$y=x$$ We'll find points of intersection between the curves for limits $$x^2=x\Rightarrow x^2-x=0$$ $$x(x-1)=0$$ $$\Rightarrow x=0,x=1$$ $$\text{For }x=0,y=0$$ $$\text{For }x=1,y=1$$ So the points of intersection are (0,0) and (1,1)

Let's do this with vertical strip: $${\int }_0^1{\int }_{x^2}^{x}xy(x+y)dxdy$$ $$={\int }_0^1{\int }_{x^2}^x(x^{2}y+x{y}^2)dxdy$$ $$={\int }_0^1{[\frac{x^2{y}^2}{2}+\frac{x{y}^3}{3}]}_{x^2}^x$$ $$={\int }_0^1[\frac{x^4}{2}+\frac{x^4}{3}-\frac{x^6}{2}-\frac{x^7}{3}]dx$$ $$={[\frac{x^5}{10}-\frac{x^7}{14}+\frac{x^5}{15}-\frac{x^8}{24}]}_0^1$$ $$=\frac{1}{8}-\frac{1}{14}+\frac{1}{15}-\frac{1}{24}$$ $$=\boxed{{\displaystyle \frac{3}{56}}}$$