# Limits of integration for triple integrals

Suppose you have the following problem: Express the mass of a solid tetrahedron T, with vertices $$(0,0,0)$$, $$(1,0,0)$$, $$(1,4,0)$$, and $$(1,0,3)$$ and density function $$p(x,y,z) = 6xyz$$. Graphically,

I am having trouble figuring out the limits of integration. I am supposed to integrate with respect to z, y, and x in that order ($$dz, dy, dx)$$

So far, I have figured out that the lower limit on $$z$$ is $$0$$, and the lower limit on $$y$$ is also $$0$$. How would one go about calculating the upper limit on z and y as well as the limits for $$x$$?

$$Mass = \int_{?_3}^{?_4}\int_0^{?_2}\int_0^{?_1} 6xyz \space dz \space dy\space dx$$

I believe there is another question similar to this, but it does not explain how to calculate the plane boundaries, which is what I'm having trouble with.

Sorry for the crappy graph.

Thanks.

• I guess it should be $\rho(x,y\color{red}{,z})=xyz$. – Jack D'Aurizio Oct 27 '18 at 9:48

The body is located between $$0$$ and $$1$$ in $$x$$-direction, hence $$\int_0^1\ldots\ dx.$$ Now for each $$x\in[0,1]$$ you get the section of the body with the plane at $$x$$ that is parallel to $$yz$$-plane. For $$x=1$$ you get the largest triangle, for $$x=0$$ you get one point (the origin), for $$x$$ somewhere in between you get a smaller proportional (similar) triangle (red in the picture below). Now the red triangle is located in $$y$$-direction between $$0$$ and some largest possible $$y(x)$$ - it is the integration limits for $$dy$$. At last, for each $$y$$ in the red triangle you can choose the integration limits for $$z$$ - from $$0$$ to the largest possible $$z$$ in the green segment. They may depend on both $$x$$ and $$y$$.