Volume between planes

Using geometric approach, the volumes bound by the two planes in the first quadrants are actually triangular-based pyramids with the same base (x-y plane) but different heights (different z-values). The triangular base is a right triangle with 6 units on the x-axis and 5 units on the y-axis, and the height difference is 5-2 = 3 units on the z-axis.

Thus, by applying the volume formula of pyramid, we also get:

V =( $\frac{1}{3}$ ) $B\times H$ = ( $\frac{1}{3}$ )( $\frac{1}{2}$ )( $5 \times 6$ )(5 - 2) = 15.

First find the cartesian equations of the two planes in the form $z=ax+by+c$ . Starting with $L_1$ and substituting the coordinates into this equation, we have:

$0=6a+c$

$0=5b+c$

$2=c$

Hence $a=\frac {-1} {3}$ , $b=\frac {-2} {5}$ and $c=2$ , and $L_1=\frac { -1 }{ 3 } x-\frac { 2 }{ 5 } y+2$ .

Repeating for $L_2$ ,

$a=\frac {-5} {6}$ , $b=-1$ and $c=5$ , and $L_2=\frac { -5 }{ 6 } x-y+5$ .

Since $L_2$ is above $L_1$ in the first quadrant, the required volume will be given by a double integral of the form $\iint _{ }^{ }{ (L_2 - L_1)dydx }$ .

To find the terminals, we need the equation of the line at which $L_1$ and $L_2$ intersect;

Since it has $x$ and $y$ intercepts at $x=6$ and $y=5$ , $\frac { x }{ 6 } +\frac { y }{ 5 } =1$

Hence $y=\frac { -5 }{ 6 } x+5$ .

The region is therefore defined by the inequalities $0\le y\le \frac { -5 }{ 6 } x+5$ and $0\le x\le 6$ .

$\therefore \quad Volume=\int _{ 0 }^{ 6 }{ \int _{ 0 }^{ \frac { -5 }{ 6 } x+5 }{ (L_ 2 - L_ 1)dydx } } \\ =\int _{ 0 }^{ 6 }{ \int _{ 0 }^{ \frac { -5 }{ 6 } x+5 }{ (\frac { -1 }{ 2 } x-\frac { 3 }{ 5 } y+3)dydx } } \\ =\int _{ 0 }^{ 6 }{ \frac { 5 }{ 24 } { (x-6) }^{ 2 }dx } \\ =\boxed { 15 }$ .