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

\iint_{x^{2}+y^{2}\leq 4} (4-x^{2}-y^{2}) dxdy

= \int_{0}^{2\pi} \int_{0}^{2} (4-r^{2})r dr d\theta

= 2\pi \int_{0}^{2} (4r-r^{3})dr

= 8\pi

Question 4. I \subseteq \mathbb{R} is a real interval, calculate the maximum value of

\int_{I} (1-x^{2}) dx.

Solution.

To calculate the maximum value of the integration, the maximal interval I=[-1,1]. Therefore, the maximum value of the integration is

\int_{-1}^{1} (1-x^{2}) dx = \frac{4}{3}.

Qustion 5. Calculate the multiple integration

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} e^{x^{2}+y^{2}} dy dx.

Solution.

Method (i).  Use the polar coordinate.

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} e^{x^{2}+y^{2}} dydx

= \int_{0}^{\pi/2} \int_{0}^{1} e^{r^{2}} r dr d\theta

= \frac{\pi}{2} \int_{0}^{1} e^{r^{2}} r dr

= \frac{\pi}{2} (\frac{e^{r^{2}}}{2}) |_{r=0}^{r=1}

= \frac{\pi}{4}(e-1).

Method (ii). Make the substitution y=sx, then dy=x ds.

The region is 0\leq x \leq 1 and 0\leq s \leq \sqrt{1-x^{2}}/x.

That is equivalent to 0 \leq s \leq \infty and 0 \leq x \leq 1/\sqrt{1+s^{2}}.

The integration is

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}/x} e^{x^{2}+s^{2}x^{2}} xds dx

= \int_{0}^{\infty} \int_{0}^{1/\sqrt{1+s^{2}}} e^{(1+s^{2})x^{2}} x dx ds

= \int_{0}^{\infty} (\frac{1}{2(1+s^{2})} e^{(1+s^{2})x^{2}} |_{x=0}^{x=1/\sqrt{1+s^{2}}}) ds

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

= \frac{e-1}{2} \arctan s|_{s=0}^{s=\infty}

= \frac{\pi}{4} (e-1).

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