1
$\begingroup$

Evaluate triple integral bounded by a solid $S$ which is a tetrahedron defined by vertices $(0,0,0)$, $(0,0,1)$, $(1,1,0)$ and $(-1,1,0)$.

$$\iiint y^2\, dV$$

So, I tried to find the plane that contains the tree vertices except $(0,0,0)$ and with that I was planning to integrate the solid as I was doing with regions like that. But the plane doesn't have a coordinate x and so I'm in trouble at the integration. In the end, I have found the one set of limits:

$ S= {{(x,y,z): 0<z<1 , 0< z <2-y , -y<x<-y}} $

$\endgroup$
1
  • $\begingroup$ I have removed the image and LaTeX-ified this for you. We can use LaTeX on this site, and it makes life quite a bit easier than managing external links :) $\endgroup$
    – Emily
    Jun 6 '16 at 15:08
1
$\begingroup$

The plane that contains all the vertices except $0,0,1)$ is easy to find; it is the XY plane. Now if we slice the tetrahedron by a plane at $z = z_0$ where $0 \leq z_0 \leq 1$ that slice is a right triangle with vertices $$(0,0,0), (1-z_0,1-z_0,0), (z_0-1,1-z_0,0) $$ so the limits of integration to use to deal with such a plane is $$ \int_{y=0}^{1-z} \int_{x=-y}^y dx \,dy $$ And the volume integral you want is $$ \int_{z=0}^1\int_{y=0}^{1-z} \int_{x=-y}^y y^2\,dx \,dy \, dz $$ SPOILER ALERT Don't read on if you want to do the integral yourself. $$ \int_{z=0}^1\int_{y=0}^{1-z} \int_{x=-y}^y y^2\,dx \,dy \, dz = \int_{z=0}^1\int_{y=0}^{1-z} \left[ xy^2 \right]_{x=-y}^{x=y} \,dy \, dz \\ = \int_{z=0}^1\int_{y=0}^{1-z} 2y^3\,dy \, dz = \int_{z=0}^1 \left[ \frac{y^4}{2}\right]_{y=0}^{1-z}\\ = \frac12\int_{z=0}^1 \left( 1-z \right)^4 \, dz=\frac{1}{10} $$

$\endgroup$
1
  • $\begingroup$ Thank you Mark! I found the same thing that you, i was editing to include that information. $\endgroup$
    – JJWho
    Jun 6 '16 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.