Question

\[\large \int_0^{3/2} \int_{x/\sqrt 3}^{\sqrt{2x - x^2}} \frac{dydx}{\sqrt{x^2 + y^2}}\]

help?

Answer 1

The region is between a circle and a line, apparently.

Answer 2

lol no graphing :p just find the integral

Answer 3

What? No graphing? o_O
I'd convert to polar coordinates but apparently that's thrown out of the window.

Answer 4

\(y=x\sec{\theta}\) would be the substitution I would use but the bounds are horrible.

Answer 5

that's what i did too and got \[\ln (\sec \theta + \tan \theta)\]

Answer 6

I meant \(y=x \tan\theta\).
Trying it anyway...
\[y=\sqrt{2x-x^2}\Rightarrow \theta=\tan^{-1}\left ( \frac{\sqrt{2x-x^2}}{x}\right )=\alpha\\
y=\frac{x}{\sqrt3} \Rightarrow \theta=\frac{\pi}{6}\\
dy=x\ \sec^2\theta d\theta\\
\Large \int_0^{3/2}\int_{\pi/6}^\alpha \frac{x \sec^2\theta d\theta dx}{x \sec\theta}\]

Answer 7

?!

Answer 8

Huh? Which part is bugging you?

Answer 9

the substitution of theta thingy...i dont know that

Answer 10

The bounds, you mean?

Answer 11

yeah

Answer 12

First integrate the inner part and then the outer part

\[\large \int\limits_0^{3/2} [{ \int\limits_{x/\sqrt 3}^{\sqrt{2x - x^2}} \frac{dy}{\sqrt{x^2 + y^2}}}]dx\]

First integrate the part in the brackets keeping y as constant

Answer 13

Sorry I meant keeping x as constant

Answer 14

uhh i think i got a weid answer for that @_@ with ln and such

Answer 15

\[\int\limits \frac{1}{\sqrt{x^2+y^2}} \, dy\]
Let
\[y=x \tan (u) ~~and ~~dy=x \sec ^2(u) d u\]
\[u=\tan^{-1}(\frac{y}{x})\]
\[\text{Then}\sqrt{x^2+y^2}\text{ =}\sqrt{x^2 \tan ^2(u)+x^2}\text{ = }x \sec (u)\]
\[\int\limits \sec (u) \, du\]
\[[\ln (\tan (u)+\sec (u))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]

Answer 16

Then after you do upper limit - lower limit, integrate with dx with the limits 0 to 3/2

Answer 17

uhmm yep i got that...then i dont know how to integrate it

Answer 18

@lgba, make sure you substitute back u = arctan(y/x) before taking upper limit-lower limit

Answer 19

Contiuing from @.Sam. 's work
so basically you get

\[[\ln (\tan (\arctan(y/x))+\sec (\arctan(y/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
\[[\ln (y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
\[[\ln (y/x)+(\sqrt{x^2+y^2}/x)]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]

Answer 20

Now take upper limit - lower limt

Answer 21

\[\small \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]
lol

Answer 22

\[\tiny \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]

Answer 23

Well sorry I messed up the bracket's earlier. It should be
\[[\ln ((y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
\[[\ln ((y/x)+\frac{\sqrt{x^2+y^2}}{x} )]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]

Answer 24

wow those are small @.Sam.

Answer 25

wolfram doesn't know how to integrate this, lol

Answer 26

\[\small \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{x^2+(\sqrt{2x-x^2})^2}}{x} )-[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]\]

Answer 27

darn my teacher -_-

Answer 28

\[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2x}}}{x} )-[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]\]
\[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )-[\ln ((\frac{1}{\sqrt{3}})+2 )]\]
So now, it's upto you to integrate
\[\int\limits_{0}^{3/2}\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx-\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx\]

Answer 29

ugh can i still quit learning this?

Answer 30

you just succeeded i hurting my poor brain :p

Answer 31

Try wolfram :P The second integral is easy. I have no idea on first part :P

Answer 32

@.Sam. said wolfram cant do it

Answer 33

@jazy, what was that ??

Answer 34

@lgbasallote , wolfram works :p
http://www.wolframalpha.com/input/?i=integral+of+ln%28%28sqrt%282x-x^2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx

Answer 35

First part is solved by wolfram alpha. Take upper limit - lower limit for the first.

The second part of integral is a cake walk.

But I would still be interested for the technique of solving http://www.wolframalpha.com/input/?i=integral+of+ln%28%28sqrt%282x-x%5E2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx

Answer 36

ugh enough =_=