Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
The region is between a circle and a line, apparently.
lol no graphing :p just find the integral
What? No graphing? o_O I'd convert to polar coordinates but apparently that's thrown out of the window.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\(y=x\sec{\theta}\) would be the substitution I would use but the bounds are horrible.
that's what i did too and got \[\ln (\sec \theta + \tan \theta)\]
I meant \(y=x \tan\theta\). Trying it anyway... \[y=\sqrt{2x-x^2}\Rightarrow \theta=\tan^{-1}\left ( \frac{\sqrt{2x-x^2}}{x}\right )=\alpha\\ y=\frac{x}{\sqrt3} \Rightarrow \theta=\frac{\pi}{6}\\ dy=x\ \sec^2\theta d\theta\\ \Large \int_0^{3/2}\int_{\pi/6}^\alpha \frac{x \sec^2\theta d\theta dx}{x \sec\theta}\]
?!
Huh? Which part is bugging you?
the substitution of theta thingy...i dont know that
The bounds, you mean?
yeah
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
Sorry I meant keeping x as constant
uhh i think i got a weid answer for that @_@ with ln and such
\[\int\limits \frac{1}{\sqrt{x^2+y^2}} \, dy\] Let \[y=x \tan (u) ~~and ~~dy=x \sec ^2(u) d u\] \[u=\tan^{-1}(\frac{y}{x})\] \[\text{Then}\sqrt{x^2+y^2}\text{ =}\sqrt{x^2 \tan ^2(u)+x^2}\text{ = }x \sec (u)\] \[\int\limits \sec (u) \, du\] \[[\ln (\tan (u)+\sec (u))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
Then after you do upper limit - lower limit, integrate with dx with the limits 0 to 3/2
uhmm yep i got that...then i dont know how to integrate it
@lgba, make sure you substitute back u = arctan(y/x) before taking upper limit-lower limit
Contiuing from @.Sam. 's work so basically you get \[[\ln (\tan (\arctan(y/x))+\sec (\arctan(y/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln (y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln (y/x)+(\sqrt{x^2+y^2}/x)]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
Now take upper limit - lower limt
\[\small \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\] lol
\[\tiny \log \left(\tan \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)+\sec \left(\tan ^{-1}\left(\frac{\sqrt{2 x-x^2}}{x}\right)\right)-\log \left(\tan \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{\frac{x}{\sqrt{3}}}{x}\right)\right)\right)\right)\right)\]
Well sorry I messed up the bracket's earlier. It should be \[[\ln ((y/x)+\sec (arcsec(\sqrt{x^2+y^2}/x)))]_{x/\sqrt3}^{\sqrt{2x-x^2}}\] \[[\ln ((y/x)+\frac{\sqrt{x^2+y^2}}{x} )]_{x/\sqrt3}^{\sqrt{2x-x^2}}\]
wow those are small @.Sam.
wolfram doesn't know how to integrate this, lol
\[\small \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{x^2+(\sqrt{2x-x^2})^2}}{x} )-[\ln (\frac{x/\sqrt3}{x})+\frac{\sqrt{x^2+(x/\sqrt3)^2}}{x} )]\]
darn my teacher -_-
\[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2x}}}{x} )-[\ln ((\frac{1}{\sqrt{3}})+\frac{\sqrt{4x^2}}{x} )]\] \[ \ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )-[\ln ((\frac{1}{\sqrt{3}})+2 )]\] So now, it's upto you to integrate \[\int\limits_{0}^{3/2}\ln ((\sqrt{2x-x^2}/x)+\frac{\sqrt{{2}}}{{\sqrt{x}}} )dx-\int\limits_{0}^{3/2}\ln ((\frac{1}{\sqrt{3}})+2 )dx\]
ugh can i still quit learning this?
you just succeeded i hurting my poor brain :p
Try wolfram :P The second integral is easy. I have no idea on first part :P
@.Sam. said wolfram cant do it
@jazy, what was that ??
@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
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
ugh enough =_=

Not the answer you are looking for?

Search for more explanations.

Ask your own question