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Find the volume of $\iint_{V}\int x dV$ when V is volume between $x=0,y=0,z=0,6x+3y+z=6$

So we are only looking on the positive side of XYZ axis, and $6x+3y+z=6\Rightarrow z=6-6x-3y$

So for $x=0$ and $y=0$ $z=6$ and decreasing so the limit of integration for z is from 0 to 6.

for x it is from $0 to 6-6x$ and for y it is from 0 to $ 6-3y$? (taking $x=0$ and $y=0$ each time?)

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  • $\begingroup$ Do you not mean $x_i\geq0$ for your 3 variables? Else we're integrating between a point and a plane which is rather interesting. $\endgroup$
    – K.Power
    Apr 11, 2016 at 11:21
  • $\begingroup$ @K.Power $x=0$ mean the YZ plan and so on, sorry if it is not well written, I will edit it $\endgroup$
    – gbox
    Apr 11, 2016 at 11:26

1 Answer 1

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Limits for $z$ are $0$ and $6-6x-3y$, for $y$ they are $0$ and $2-2x$ and for $x$ they are $0$ and $1$. This is just one way to set up the integral, basically you can choose every possible order of $x$, $y$ and $z$. You just set all enclosed variables to zero for determining the integration borders for each variable.

So the integral is given by

$$\int\int\int_{V}x\,dV = \int_{0}^{1}\int_{0}^{2-2x}\int_{0}^{6-6x-3y}x\;dz\,dy\,dx$$

First integration over $z$ yields

$$\int_{0}^{1}\int_{0}^{2-2x}\left(6x-6x^2-3xy\right)dy\,dx$$

Then integration over $y$ gives

$$\int_{0}^{1}\left[6xy-6x^2y-\frac{3}{2}xy^2\right]_0^{2-2x}dx=12\int_{0}^{1}\left(\frac{1}{2}x^3-x^2+\frac{1}{2}x\right)dx$$

Finally integrating over $x$ results in

$$12\left[\frac{1}{8}x^4-\frac{1}{3}x^3+\frac{1}{4}x^2\right]_0^1 = \frac{1}{2}$$

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