How to evaluate the triple integral where is the solid tetrahedon with vertices ?
thankyou!
Here's a similar problem although the points in that post aren't showing up but still it's the same: you first have to calculate a normal to the plane you're integrating under, then derive the equation of the plane, then integrate over the appropriate region in the x-y plane.
http://www.mathhelpforum.com/math-he...trahedron.html
Maybe we should work this one through since it's been asked a few times. Using the plot below which is a little hard to visualize unless you can rotate it, I'll calculate two vectors: One goes from the point at (0,3,0) to (0,0,2) and the other goes from (8,0,0) to (0,3,0):
$\displaystyle v_1=<0,3,0>-<8,0,0>=<-8,3,0>$
$\displaystyle v_2=<0,0,2>-<0,3,0>=<0,-3,2>$
In order to calculate a normal to the blue surface, I calculate the cross-product of these two vectors which I've indicated as the line eminating from the origin.
$\displaystyle \left|\begin{array}{ccc} i & j & k \\
-8 & 3 & 0 \\
0 & -3 & 2
\end{array}\right|$
$\displaystyle n=6i+16j+24k$
I can now derive the equation of the blue surface using the equation of the plane passing through the point (0,3,0) and the normal vector:
$\displaystyle n_x(x-p_x)+n_y(y-p_y)+n_z(z-p_z)=0$
$\displaystyle 6(x-0)+16(y-3)+24(z-0)=0$
or: $\displaystyle z=f(x,y)=\frac{-6x-16(y-3)}{24}$
That line in the back is $\displaystyle y=-3/8x+3$ right?
So then the integral is:
$\displaystyle V=\int_0^8\int_0^{-3/8x+3}\int_0^{f(x,y)} xydzdydx$
Pretty sure anyway.