Convert triangle vertices to double integral polar coordiantes

ramses07
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Homework Statement



integrate

f(x,y) = sqrt(x^2+y^2)

over triangle with vertices (0,0) (0,sqrt2) (sqrt 2, sqrt 2)

Homework Equations



x= rcosO, y = rsinO

x^2+y^2=r^2

The Attempt at a Solution



im supposed to use a double integral converted to polar coordinates,
so i used the bounds int. 0 to pi/4 int. 0 to sqrt 2 sec (r^2) drdO

are these the correct bounds? because i can't seem to find the answer.
 
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I think your bounds for theta should be pi/4 to pi/2
 
how do you know what the bounds are?
 
because the line y=x cuts it at a 45 degree angle and x=0 goes up to 90 so it goes from 45 to 90
like cutting a wedge out of a circle , but the radius is different , i think your bounds for the radius are correct.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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