integration...should i use polar transformation?

**Sambit** · October 21st 2010, 07:33 AM

Evaluate $\iint\limits_R \frac{1}{(1 + x^2 + y^2)^\frac{3}{2}} dx dy$
where $R$ is the region bounded by the straight lines $y=0 , x=1 , y=x$

if i use polar transformation, what will be the ranges of integration? or is there any way other than polar transformation?

**Ackbeet** · October 21st 2010, 07:45 AM

I think I'd stay in cartesian coordinates for this one. What is your integral, including limits?

**Sambit** · October 21st 2010, 07:56 AM

writing in terms of limits, the integration becomes $\int_0^1\int_0^x \frac{1}{(1 + x^2 + y^2)^\frac{3}{2}} dy dx$

how to evaluate this?

**Ackbeet** · October 21st 2010, 08:00 AM

Right. So the $y$ indefinite integral is of the form

$\displaystyle\int\frac{dy}{(a^{2}+y^{2})^{3/2}}.$

How would you compute that antiderivative?

**Sambit** · October 21st 2010, 08:20 AM

we substitute $y = a \tan\theta$ , which gives the final value of the above integral as $\frac{1}{a^2}\sin y$ , ie, $\frac{sin x}{1+x^2}$ . right?? then what?

**Ackbeet** · October 21st 2010, 08:23 AM

I think you may have skipped too many steps there. I agree with the substitution, but I get a different result. How do the integrand and differential change with your trig substitution?

**Sambit** · October 21st 2010, 08:31 AM

we substitute $y = a \tan\theta$ , which gives the final value of the above integral as $\frac{1}{a^2} \frac{y}{\sqrt{y^2+a^2}}$ . ok? that is $\frac{x}{(1+x^2)(\sqrt{1+2x^2})}$ . right??

**Ackbeet** · October 21st 2010, 08:35 AM

The final result is correct. So, now what? You've got

$\displaystyle\int_{0}^{1}\frac{x}{(1+x^2)(\sqrt{1+ 2x^2})}\,dx.$

Ideas?

**Sambit** · October 21st 2010, 08:44 AM

here we substitute $x^2 = z$ then $1 + 2z = t^2$ this gives us the final result as $\arctan\sqrt{2x^2 + 1}$ . am i right?

**Ackbeet** · October 21st 2010, 08:46 AM

Not sure I'm following your steps here. The final result is incorrect, actually. It should be arctan.

**Sambit** · October 21st 2010, 08:49 AM

ya...sorry..it will be tan inverse....just a little typing mistake.....sorry for that

i have made the correction in the previous post. so the answer comes to be $\arctan \sqrt{3} = \frac{\pi}{3}$

**Ackbeet** · October 21st 2010, 09:02 AM

Don't forget the lower limit.

**Sambit** · October 21st 2010, 09:08 AM

oooops....so much silly mistakes...

... yeah. the answer is then $\frac{\pi}{3}-\frac{\pi}{4} = \frac{\pi}{12}$ .....

many many thanks:d

**Ackbeet** · October 21st 2010, 09:19 AM

There you go. Nice work.

One way to avoid silly mistakes is to pretend that someone is holding a gun to your temple, and if you make a mistake, he's going to pull the trigger. Or, you could work at a high-energy physics lab, with all those 10,000V wires running around. If you make a mistake, you're dead.

So there are some nice, cheerful ways to avoid mistakes.

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