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I am stuck on the following problem.

Let $R = [0,1] \times [0,1]$. Find the volume of the region above $R$ and below the plane which passes through the 3 points $(0,0,1), (1,0,8), (0,1,2)$.

I know $R$ is a square with area 1 cubit unit, and the equation of plane can be determined by using the 3 points. What I am confused on is finding the volume once I have both of those parts, since the plane equation would have 3 variables $(x,y,z)$ and R would only have 2?

Thank you in advance

Ottavio
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mathjohnn
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2 Answers2

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As I found, your plane is $z=7x+y+1$. So your volume is $$\int\limits_{0}^{1}\int\limits_{0}^{1}\int\limits_{0}^{7x+y+1}dxdydz = \int_\limits{0}^{1}\int_\limits{0}^{1}(7x+y+1)dxdy$$

zkutch
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  • Thank you. I didn't know the plane's equation was supposed to be a bound for the third integral. I actually didn't know there was supposed to be a third integral haha. This helped alot. – mathjohnn Jul 16 '20 at 22:20
  • @mathjohnn: It is possible to do it either as Basco did in their answer, as a double integral, treating the height of the plane $z = 7x+y+1$ as the integrand over $x$ and $y$, or as zkutch does in this answer, as a triple integral, treating the height of the plane as the upper limit of integration and using $1$ as the integrand. The two approaches are ultimately equivalent: As zkutch also shows, evaluating the inner ($z$) integral in the triple integration directly produces the double integration. – Brian Tung Jul 16 '20 at 22:32
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Imagine the conventional Rectangular Cartesian coordinate system and call the axes x,y and z as you did in your question. Consider that the plane of the base is x-y and the axis z is perpendicular to it. Once you have the equation of the plane (which is something alike to $0= ax+by+cz+d$, where a,b,c and d are reals) you can place a small element of volume of size $dV=dx dy z(x,y)$ where $z(x,y)$ is the height of $dV$ from the plane $x-y$ up to the plane above, can be found from the equation for the plane. Therefore, instead of integrating 3 times from 0 to 1 in $x$ and $y$ and from zero to $z(x,y)$ in $z$ you can do it only twice:

$V=\int^1_0 \int^1_0 z(x,y) dx dy$

Basco
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