Density, Volume and Slicing Problem

[Eonfluxx]

Homework Statement

A compressible liquid has density which varies with height. AT the level of h meters above the bottom, the density is 40(5 - h) kg/m³

a) The liquid is put in the containers below. The cross sections of the container are isosceles triangles. It has straight sides and looks like a triangular prism. How many kg will it hold when placed as shown on the left, resting on one triangular side?​

^There is a picture of the object.

Homework Equations

Mass = Volume times Density

The Attempt at a Solution

My theory is that since M = V*D, and you're given the density, should I slice and solve for volume, then evaluate the integral of volume and the given integral for density then multiply the results to get the mass?
\int^{4}_{0}2.5 dh and 40\int^{4}_{0}5-h dh

I got to the first integral by taking the volume of the first slice (triangular prism):
\sum\frac{1}{2}b*l*\Delta h \rightarrow \sum\frac{1}{2}2*2.5\Delta h \rightarrow lim as \Delta h \rightharpoonup\int^{4}_{0}2.5 dh

To find the mass, according to the equation M=VD, should I solve both integrals , one being V and one being D, then multiply?

 

[Eonfluxx]

Ah... It's starting to make more sense now, thanks. So since density is varying with the height we make it all one integral rather than separate ones. Part B is the same question except with the 4m on the ground on its point with the rectangle side up. I figured once I'd managed A, B would be simple. Thanks!

 

[cronxeh]

After some deep thought, kicking a dog, and watching The Core in 1080p quality, I've decided to set up a triple integral.

2*int 40*(5-z) dy dx dz , y=0..-2.291*x+2.291, x=0..1, z=0..4

http://www.wolframalpha.com/input/?i=2*int+40*(5-z)+dy+dx+dz+,+y=0..-2.291*x+2.291,+x=0..1,+z=0..4"

It looks like 1099.68 kg is the correct answer

The original idea to integrate using int(2.291*40*(5-h),h,0,4) = 1099.68