# Evaluate \int_0^2\int_0^x \sqrt {x^2+y^2}dydx+ \int_2^{2\sqrt 2} \int_0^{\sqrt{8-x^2}} \sqrt...

## Question:

Evaluate {eq}\displaystyle \int_0^2\int_0^x \sqrt {x^2+y^2}dydx+ \int_2^{2\sqrt 2} \int_0^{\sqrt{8-x^2}} \sqrt {x^2+y^2} dydx {/eq}

## Double Integral in Polar Form:

The Cartesian system navigating a point based on it's distance from x,y and sometimes z axis.

In polar form,there are generally two parameters for navigating to a point {eq}\displaystyle r\:and\:\theta {/eq}

r is the distance to that co ordinate point .{eq}\displaystyle \theta {/eq} represents the angle that aforementioned theoretical vector would make with x-axis.

Representation of Cartesian form into polar form is as follows:

{eq}\displaystyle \int \:\int _D\:\:f\left(x,y\right)dxdy=\int _{\alpha }^{\beta }\int _{h_1\left(\theta \right)}^{h_2\left(\theta \right)}f\left(r\:cos\:\theta ,\:r\:sin\:\theta \right)\:\:\:r\:dr\:d\theta {/eq}

Given integral is

{eq}\displaystyle I=\int_0^2\int_0^x \sqrt {x^2+y^2}dydx+ \int_2^{2\sqrt 2} \int_0^{\sqrt{8-x^2}} \sqrt {x^2+y^2} dydx\\ \displaystyle I=I_1+I_2 {/eq}

Let's convert it into polar form ,as

{eq}\displaystyle x=rcos\:\theta \:;\:y=rsin\:\theta \\ \displaystyle dxdy=r\:dr\:d\theta {/eq}

Hence,

For {eq}\displaystyle I_1 {/eq} part,

Substitute, {eq}\displaystyle x^2+y^2=r^2\\ \displaystyle y=x\:gives,\:r\:sin\:\theta =r\:cos\:\theta ==>tan\:\:\theta =1\:i.e\:\theta =\frac{\pi }{4}\\ \displaystyle \therefore \:0\:\le \theta \:\le \:\frac{\pi \:}{4}\\ \displaystyle Also,x=2\:gives,r\:cos\:\theta =2==>rcos\left(\frac{\pi }{4}\right)=2==>r=2\sqrt{2}\\ \displaystyle \therefore \:0\le r\le \:2\sqrt{2} {/eq}

For {eq}\displaystyle I_2 {/eq} part,

Substitute, {eq}\displaystyle x^2+y^2=r^2\\ \displaystyle y=\sqrt{8-x^2}\:i.e\:\:x^2+y^2=8\:gives\:,\:r^2=8\\ \displaystyle \therefore \:0\:\le r\:\le \sqrt{8}\\ \displaystyle Also,\:x=2\:gives,r\:cos\:\theta \:=2==>\sqrt{8}cos\left(\theta \right)=2==>\theta =cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}\:\:and\:x=2\sqrt{2}\:gives,\:\theta =0\\ \displaystyle \therefore \:\:\frac{\pi \:}{4}\le \:\theta \:\le 0\: {/eq}

Hence, polar form integral becomes,

{eq}\displaystyle I=\int _0^{2\sqrt{2}}\int _0^{\frac{\pi }{4}}\:\sqrt{r^2}r\:dr\:d\theta +\:\int _{\frac{\pi }{4}}^0\:\int _0^{2\sqrt{2}}\:\sqrt{r^2}r\:dr\:d\theta \:\\ \displaystyle I=\int _0^{2\sqrt{2}}\int _0^{\frac{\pi }{4}}\:r^2dr\:d\theta -\int _0^{\frac{\pi \:}{4}}\:\int _0^{2\sqrt{2}}\:r^2\:dr\:d\theta \:\\ \displaystyle I=0 {/eq}