asked 2014-07-24 10:14:06 -0600
I just want to know if I am doing this correctly. On the worksheet from today, number 6 asks: Set up an integral representing the volume under the plane $x+2y+z=2$ and in the first octant.
I came up with the following: $$\int_0^2\int_0^\frac{2-x-z}{2}\int_0^{2-2y-z} dxdydz$$
am I doing this right or am I completely lost?
answered 2014-07-24 13:27:52 -0600
I did mine with my integral in the order $\delta x\delta y\delta z$ like yours, Anonymous, and I got my final triple integral to be $$\int_{0}^{2}\int_{0}^{\frac{2-z}{2}}\int_{0}^{2-2y-z}$$. We have done the same thing except for the $\delta y$ integral. I believe that for the $\delta y$ integral we set in the original equation $x = 0$ and solve for $y$, which gives us $2y + z = 2$ or, when we solve for $y$, $y = \frac {2-z}{2}$. Because we are solving for the volume of the plane in the first octant which means all $x, y,$ and $z$ values must be positive, we know that the bounds for $y$ will be $0\leq y\leq \frac {2-z}{2}$.
This is basically what Tiffany said, but because we both did our integrals in the same order, I thought it might be helpful. Best of luck on the quiz tomorrow!
answered 2014-07-24 10:26:06 -0600
For this one my group took the $\delta z$ integral and set it as $$ \int _0 ^ {2-2y-x} $$ and then to get the $\delta y $ integral we set z=0 and solved for y to get $$\int _0 ^{ 1 - \frac x2} $$ then your x integral is $$ \int _0 ^2$$ with it going $\delta z \delta y \delta x $
Asked: 2014-07-24 10:14:06 -0600
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Last updated: Jul 24 '14
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