How do you find the volume of the solid bounded by the coordinate planes and the plane #8x + 6y + z = 6#?

Question

William Abdalla · Accepted Answer

#3/4#

use double integral (or triple if you like, i'll just do double as triple here is just extra unnecessary formality)

first we need to find the volume in question.

#8x + 6y + z = 6#

it hits the x,y,z axes as follows

#y,z = 0, x = 3/4#

#x,z = 0, y = 1#

#x,y = 0, z = 6#

so we can start with a drawing!!

so it's just a case now of finding the integration limits for this double integral #int int \ z(x,y) dA = int int \ 6 - 8x - 6y \ dx \ dy#

in the x-y plane we have

so either of these is fine

# int_{y=0}^{1} int_{x=0}^{3/4 (1-y)} \ 6 - 8x - 6y \ dx \ dy#

# int_{x=0}^{3/4} int_{y=0}^{1 - 4/3 x} \ 6 - 8x - 6y \ dy \ dx#

Either way, I get answer 3/4

Isabella Ackerman · Answer

undefined To find the volume of the solid bounded by the coordinate planes and the plane $8x + 6y + z = 6$, we can use the method of triple integration with appropriate limits.

The volume $V$ can be expressed as a triple integral over the region $D$ in the $xyz$-space bounded by the planes $x = 0$, $y = 0$, $z = 0$, and $8x + 6y + z = 6$:

$$ V = \iiint_D dV $$

To set up the limits of integration, we first find the intersection points of the plane with the coordinate planes:

When $x = 0$, $y = 0$, $z = 0$, and $8x + 6y + z = 6$, we get the points $(0, 0, 6)$, $(0, 1, 0)$, and $(\frac{3}{4}, 0, 0)$, respectively.

Now, the limits of integration for $x$, $y$, and $z$ are as follows:

For $x$: $0 \leq x \leq \frac{3}{4}$  
For $y$: $0 \leq y \leq 1 - \frac{4}{3}x$  
For $z$: $0 \leq z \leq 6 - 8x - 6y$

So, the volume $V$ can be computed by evaluating the triple integral:

$$ V = \int_0^{\frac{3}{4}} \int_0^{1 - \frac{4}{3}x} \int_0^{6 - 8x - 6y} dz \, dy \, dx $$