Find the volume of the solid bound by the three coordinate planes

[chwala]

Homework Statement

See attached. This was straightforward. I am looking at an alternative approach hence sharing the post for insight.

Relevant Equations

Double integration .

Also,

$$V=\dfrac{1}{4} \int_4^0 \left[\int_{2-0.5y}^2 (4-2x-y) dx\right] dy$$

$$V=\dfrac{1}{4} \int_4^0\left [4x-x^2-xy]^2_{2-0.5y} \right] dy$$

$$V=\dfrac{1}{4} \int_4^0 \left [(4-2y)-(4(2-0.5y)-(2-0.5y)^2-(2-0.5y)y] \right] dy$$

$$V=\dfrac{1}{4} \int_4^0 \left [(4-2y)-(2-0.5y)^2 \right] dy$$

$$V=\dfrac{1}{4} \int_4^0 \left [\dfrac {-1}{4} y^2\right] dy =\left[\dfrac{-y^3}{48}\right]_4^0=0+\dfrac{64}{48}=\dfrac{4}{3}$$

Bingo!! cheers any insight welcome.
...below is my first attempt that is clearly wrong...i corrected it my final post above... i had used wrong limits on my working...

 

[pasmith]

By the divergence theorem,$$\begin{split} \int_V 1\,dV &= \frac13 \int_V \nabla \cdot \mathbf{x}\,dV \\ &= \frac13 \int_{\partial V} \mathbf{x} \cdot \mathbf{n}\,dS \\ &= \frac13\underbrace{\int_{x=0} -x \,dy\,dz}_{=0} + \frac13\underbrace{\int_{y=0} -y \,dx\,dz}_{=0} + \frac13\underbrace{\int_{z=0} -z \,dx\,dy}_{=0} + \frac13\int_A \mathbf{x} \cdot \mathbf{n}\,dS \\ &= \frac13\int_A (2u, 4v, 1 - u - v) \cdot ((2,0,-1) \times (0,4,-1))\,du\,dv \\ &= \frac13\int_A 2(2u, 4v, 1-u-v) \cdot (2,1,4)\,du\,dv \\ &= \frac{8}{3} \int_0^1 \int_0^{1-u}1\,dv\,du \\ &= \frac{4}{3}.\end{split}$$

 

[Steve4Physics]

The plane 2x+ y + 4z = 4 intersects
- the x-axis (y=z=0) when 2x = 4, i.e. at (2, 0, 0);
- the y-axis (x=z=0) when y = 4, i.e. at (0, 4, 0);
- the z-axis (x=y=0) when 4z = 4, i.e. at (0, 0, 1).

The shape is a pyramid of height (z) 1 with a base (on xy plane) of area $\frac 12$ x 2 x 4 = 4.

The volume of a pyramid is $\frac 13$ x base-area x height = $\frac 13$ x 4 x 1 = $\frac 43$.