Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is ; How can I solve the following integral question?

$\displaystyle \int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$

Thanks in advance,

share|improve this question
1  
What have you tried so far? –  user17762 Jan 29 '11 at 15:17

4 Answers 4

up vote 7 down vote accepted

You've had some time to study this, so let's look closer at the two evident approaches: (a) conversion to polar coordinates, (b) integrate directly with a standard hyperbolic substitution.

(a) Conversion to polar coordinates: Let

$$I = \int_{1/\sqrt{2}}^1 \, \int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y \, \text{d}x.$$

If you draw a diagram of the situation you will see that

$$I = \int_0^{\pi/4} \, \int_{r=1}^{r=\sec \theta} \frac{1}{r} r \text{d}r \, \text{d} \theta = \int_0^{\pi/4} \sec \theta \, - 1 \textrm{ d} \theta $$

$$= \left[ \frac{1}{2} \log \left| \frac{1+ \sin \theta}{1- \sin \theta} \right| - \theta \right]_0^{\pi/4} = \log(1+\sqrt{2}) - \frac{\pi}{4}.$$

(b) Standard hyperbolic substitution: To evaluate the inner integral set $y= x \sinh u$ and noting that $\sinh^{-1} u = \log(u+ \sqrt{1+u^2})$ we have

$$\int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y = \left[ \log \left( \frac{y}{x} + \sqrt{ 1+ \frac{y^2}{x^2} }\right) \right]_{\sqrt{1-x^2}}^x $$ $$= \log(1+ \sqrt{2}) + \log x - \log( 1+ \sqrt{1-x^2}). \quad (1)$$

Both the logs can be integrated by parts. The first is standard, $\int \log x \textrm{ d}x = x \log x - x + C$ and the second

$$ \int \log( 1+ \sqrt{1-x^2} ) \textrm{ d}x = x \log( 1+ \sqrt{1-x^2}) -x + \sin^{-1} x + C.$$

You will need (do the integration) to note that

$$\frac{1}{\sqrt{1-x^2}} - 1 = \frac{x^2}{(1-x^2) + \sqrt{1-x^2}}.$$

And so upon integrating $(1)$ between $1/\sqrt{2}$ and $1$ you obtain

$$I = \left(1 - \frac{1}{\sqrt{2}} \right) \log(1+ \sqrt{2}) + \left[ x \log x - x\log( 1+ \sqrt{1-x^2}) - \sin^{-1} x \right]_{1/\sqrt{2}}^1$$ $$= \log(1+\sqrt{2}) - \frac{\pi}{4}.$$

share|improve this answer

Have you tried matlab? If you say:

int('1/sqrt(x^2+y^2)','y','sqrt(1-x^2)','x')

then you'll get: log(x + (2*x^2)^(1/2)) - log((1 - x^2)^(1/2) + 1) as first integral.

Then: int('log(x + (2*x^2)^(1/2)) - log((1 - x^2)^(1/2) + 1)', 'x')

Result:

x*(log(x + (2*x^2)^(1/2)) - 1) - int(log((1 - x^2)^(1/2) + 1), x)

But the last part can't be integrated symbolically, so, in matlab, you can evaluate it numerically:

`double(int('log((1 - x^2)^(1/2) + 1)', 'x',sqrt(2)/2,1))` 

which will give : ans = 0.1143, and the other part you can evaluate yourself. It should be easy to replace x with the 2 numbers. Hope this helps!

share|improve this answer
    
Thanks for your alternative answer. –  MAxcoder Jan 29 '11 at 19:41
    
Have you checked it? Is it ok? I haven't checked it, i just copied from matlab. It's quite cool. You can also try wolframalpha, it shows the steps, but it didn't work for me (computation timeout), heh, it was to tough for it. –  Firte Andrei Jan 30 '11 at 0:18

Try converting to polar coordinates, you should end out integrating $\sec(\theta) - 1$.

share|improve this answer
    
Thank you very much. –  MAxcoder Jan 29 '11 at 19:22

Hint:for the y integral, consider x to be a constant. It is a standard form that can be solved by a trigonometric substitution.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.