Triple Integral over the volume bounded by

[jegues]

Homework Statement

Evaluate the triple integral of the function $$f(x,y,z) = x$$ over the volume bounded by the surfaces

$$2x + 3y + z =6,x=0,y=0,z=0.$$

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

I sketched the volume bounded by the surfaces and set my integrals.

I question if I chose the right order to integrate because if I would have chosen to integrate the in the x direction first I wonder if it would have made things smoother.

I chose not to do this because then then top bound of my first integral would be fairly ugly, i.e.

$$x = \frac{6-3y-z}{2}$$

On the flip side, when I chose to go with the z direction first I got no fractions on my upper bound of my first integral however the integrations that come after are ugly.

What do you guys think? Did I even get my integrals correct?

Thanks again!

 

[Mark44]

Your first integral looks fine.
$$\int_{x=0}^3 \int_{y = 0}^{-2/3 x + 2} \int_{z = 0}^{6 - 2x - 3y} x~dz~dy~dx$$