Determining Volume using double integral

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Homework Statement



Find the volume of the solid in the first octant by the planes z = x + y + 1 and z = 5 - x - y

Homework Equations





The Attempt at a Solution



How would I set this up?
 
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You would set it up by drawing the region in the z=0 (i.e. xy plane) and integrating z*dx*dy over that region.
 
Are you sure it's not supposed to be dz dy dx? I went to my prof with this one, and the way he set it up as a series of triple integrals.
 
Your post title was "Determining Volume using double integral". It you want to set it up as a triple integral first then you integrate 1*dz*dx*dy over the volume. The integral of 1*dz is just z. It's the same thing. Now get started.
 
http://img16.imageshack.us/img16/1340/39161133fw1.jpg

Its not to scale though; I just need to worry about the top right quadrant.

\int^{5}_{0}\int^{x - 5}_{0}\int^{5- x - y}_{0} dz dy dx + \int^{1}_{0}\int^{-x - 1}_{0}\int^{x + y +1}_{0} dz dy dx

How off am I?
 
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Not too good I don't think. I'm not too sure what you are trying to draw, but your first step should be to figure out where the two planes intersect. Can you do that? I think that will help you picture the region.
 
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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