Finding the Volume Under a Surface Above a Triangle

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Homework Statement


find the volume under the surface z=(y)(x)^2 and above the triangle with vertices (4,0) (1,0) (2,1)



Homework Equations





The Attempt at a Solution



i got my limits of integration to be y=0 to y=1 and x=y+1 to x=4-2y. my integrand itself is (y)(x)^2dxdy

i just wanted to see if this is correct.
thanks!
 
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help please!
 
nobody ?
 
Mark44 said:
That should work.

^^^ thanks mark
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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