MA 1505 Tutorial 7: Integration of Two Variables Functions

In the tutorial 7, we will learn to calculate the integration of two variables, reverse the order of integration and polar coordinate.

The formulas of polar coordinate are $x=r \cos(\theta)$ , $y=r \sin(\theta)$ , where $r\in (0,\infty)$ and $\theta \in [0, 2\pi)$ .

$\iint_{D} f(x,y) dxdy= \iint_{D^{'}} f(r \cos \theta, r \sin \theta) r dr d\theta$

Question 1. The application of polar coordinate. Calculate the value of

$I= \int_{-\infty}^{\infty} e^{-x^{2}}dx.$

Solution.

Method (i).

$I=\int_{-\infty}^{\infty} e^{-x^{2}} dx= \int_{-\infty}^{\infty} e^{-y^{2}}dy$ .

Therefore

$I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} dxdy$

$= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^{2}} r dr d\theta$

$= 2\pi \int_{0}^{\infty} e^{-r^{2}}r dr$

$= 2\pi \frac{1}{2} e^{-r^{2}}|_{r=0}^{r=\infty}$

$= \pi$ .

Hence $I=\sqrt{\pi}$ .

Method (ii).

Since $I=\int_{-\infty}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-y^{2}}dy$ , we get

$I^{2}=\int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}-y^{2}} dy dx$

Assume y=sx, we get

$I^{2}=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}(1+s^{2})} x ds dx$

$=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-(1+s^{2})x^{2}} x dx ds$

$=4 \int_{0}^{\infty} \frac{1}{2(1+s^{2})} ds$

$=4 \cdot \frac{1}{2} \arctan s|_{s=0}^{s= \infty}$

$= \pi$

Therefore, $I=\sqrt{\pi}$

Question 2. Calculate the value of

$\lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{x}^{t} \sin{y^{2}} dy dx}{t^{4}}.$

Solution.

Method (i). Leibniz Integration Rule.

$\frac{d}{d\theta} ( \int_{a(\theta)}^{b(\theta)} f(x,\theta)dx)$

$= \int_{a(\theta)}^{b(\theta)} f_{\theta}(x,\theta) dx + f(b(\theta), \theta)\cdot b^{'}(\theta) - f(a(\theta),\theta) \cdot a^{'}(\theta)$

Here $f_{\theta}(x,\theta)$ denotes the partial derivative of $f(x, \theta)$ with respect to the variable $\theta$ .

In the question, assume $G(x,t)=\int_{x}^{t} \sin{y^{2}} dy$ .

Making use of L’Hospital Rule, we have

$\lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{x}^{t} \sin{y^{2}} dy dx}{t^{4}}$

$= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} G(x,t)dx}{t^{4}}$

$= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} G_{t}(x,t)dx+ G(t,t)\cdot 1 - G(0,t)\cdot 0}{ 4 t^{3}}$

$= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \sin{t^{2}}dx}{4t^{3}}$

$= \lim_{t\rightarrow 0^{+}} \frac{ t \sin{t^{2}}}{4t^{3}}= \frac{1}{4}$

Method (ii). Reverse the order of integration.

The integration domain is $0\leq x \leq t$ and $x \leq y \leq t$ . It is same as $0\leq y \leq t$ and $0\leq x\leq y$ .

$Answer= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{0}^{y} \sin{y^{2}} dxdy}{t^{4}}$

$= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} y \sin{y^{2}}dy}{t^{4}}$

$= \lim_{t\rightarrow 0^{+}} \frac{ t \sin{t^{2}}}{4t^{3}}$

$=\frac{1}{4}$ .

Question 3. MA1505 2010-2011 Semester 2, Question 6(b).

Let R be a region of xy-plane, find the largest possible value of the integration

$\iint_{R} (4-x^{2}-y^{2})dxdy.$

Solution.

Since we want to find the largest possible value, then we must guarantee that on the region R, the function $f(x,y)=4-x^{2}-y^{2}$ is non-negative. That means the region R is $4-x^{2}-y^{2}\geq 0$ . i.e. $x^{2}+y^{2}\leq 4$ . Therefore, we should calculate the integration