What is the double integral for a triangle with vertices (1,1),(2,4),(5,2)?

jahlin
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Homework Statement



Compute the double integral where the region is a triangle with vertices (1,1),(2,4),(5,2).(please see the attachment)

Homework Equations


double integral((16-2x-3y)1/11)da




The Attempt at a Solution


first i found the equation of the three lines :

(1,1)-->(5,2) y=x(1/4)+3/4
(1,1)-->(2,4) y=3x-2
(2,4)--->(5,2) y=-x(2/3)+16/3

and then i set up the double integral by defining the range of x and y limits.
1<x<2 , x(1/4)+3/4<y<3x-2
2<x<5, (1/4)x+3/4<y<-x(2/3)+16/3

i integrated these 2 integrals but I am not getting the right answer!
 

Attachments

  • double integral.JPG
    double integral.JPG
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Your set-up sounds correct. Show us what you did when you integrated. Maybe the problem is there.
 
if the setup sounds good then i will manage to get the right answer by myself. Thanks
 
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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