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anonymous
 4 years ago
\[\large \int_0^{3/2} \int_{x/\sqrt 3}^{\sqrt{2x  x^2}} \frac{dydx}{\sqrt{x^2 + y^2}}\]
help?
anonymous
 4 years ago
\[\large \int_0^{3/2} \int_{x/\sqrt 3}^{\sqrt{2x  x^2}} \frac{dydx}{\sqrt{x^2 + y^2}}\] help?

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blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0The region is between a circle and a line, apparently.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol no graphing :p just find the integral

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0What? No graphing? o_O I'd convert to polar coordinates but apparently that's thrown out of the window.

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0\(y=x\sec{\theta}\) would be the substitution I would use but the bounds are horrible.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's what i did too and got \[\ln (\sec \theta + \tan \theta)\]

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0I meant \(y=x \tan\theta\). Trying it anyway... \[y=\sqrt{2xx^2}\Rightarrow \theta=\tan^{1}\left ( \frac{\sqrt{2xx^2}}{x}\right )=\alpha\\ y=\frac{x}{\sqrt3} \Rightarrow \theta=\frac{\pi}{6}\\ dy=x\ \sec^2\theta d\theta\\ \Large \int_0^{3/2}\int_{\pi/6}^\alpha \frac{x \sec^2\theta d\theta dx}{x \sec\theta}\]

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0Huh? Which part is bugging you?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the substitution of theta thingy...i dont know that

blockcolder
 4 years ago
Best ResponseYou've already chosen the best response.0The bounds, you mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry I meant keeping x as constant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhh i think i got a weid answer for that @_@ with ln and such

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits \frac{1}{\sqrt{x^2+y^2}} \, dy\] Let \[y=x \tan (u) ~~and ~~dy=x \sec ^2(u) d u\] \[u=\tan^{1}(\frac{y}{x})\] \[\text{Then}\sqrt{x^2+y^2}\text{ =}\sqrt{x^2 \tan ^2(u)+x^2}\text{ = }x \sec (u)\] \[\int\limits \sec (u) \, du\] \[[\ln (\tan (u)+\sec (u))]_{x/\sqrt3}^{\sqrt{2xx^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then after you do upper limit  lower limit, integrate with dx with the limits 0 to 3/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhmm yep i got that...then i dont know how to integrate it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@lgba, make sure you substitute back u = arctan(y/x) before taking upper limitlower limit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Contiuing from @.Sam. 's work so basically you get \[[\ln (\tan (\arctan(y/x))+\sec (\arctan(y/x)))]_{x/\sqrt3}^{\sqrt{2xx^2}}\] \[[\ln (y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2xx^2}}\] \[[\ln (y/x)+(\sqrt{x^2+y^2}/x)]_{x/\sqrt3}^{\sqrt{2xx^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now take upper limit  lower limt

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1\[\small \log \left(\tan \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)+\sec \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)\right)\log \left(\tan \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\] lol

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1\[\tiny \log \left(\tan \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)+\sec \left(\tan ^{1}\left(\frac{\sqrt{2 xx^2}}{x}\right)\right)\log \left(\tan \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well sorry I messed up the bracket's earlier. It should be \[[\ln ((y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2xx^2}}\] \[[\ln ((y/x)+\frac{\sqrt{x^2+y^2}}{x} )]_{x/\sqrt3}^{\sqrt{2xx^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow those are small @.Sam.

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1wolfram doesn't know how to integrate this, lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\small \ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{x^2+(\sqrt{2xx^2})^2}}{x} )[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{{2x}}}{x} )[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]\] \[ \ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )[\ln ((\frac{1}{\sqrt{3}})+2 )]\] So now, it's upto you to integrate \[\int\limits_{0}^{3/2}\ln ((\sqrt{2xx^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ugh can i still quit learning this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you just succeeded i hurting my poor brain :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try wolfram :P The second integral is easy. I have no idea on first part :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@.Sam. said wolfram cant do it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jazy, what was that ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@lgbasallote , wolfram works :p http://www.wolframalpha.com/input/?i=integral+of+ln%28%28sqrt%282xx^2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First 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%282xx%5E2%29%29%2Fx+%2B+sqrt%282%2Fx%29%29dx
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