Let the planes be defined by $$|x|+|y|+|z|=1$$
Find the volume of the solid enclosed and the total surface area of the solid thus generated.
I am not able visualise the solid. What will it be$?$ I can get the equations of all the planes that it will generate under the given condition, but I am not able to determine the solid and then how to find the volume and it's surface area. Also, I am unsure about using integration here$?$
Any help is greatly appreciated.
On
Let $|x| = \alpha x, |y| = \beta y, |z| = \gamma z$, and for the volume consider the integral $$\alpha \beta \int_0^{\alpha} \int_0^{\frac{(1-\alpha x)}{\beta}}(1- \alpha x - \beta y) dy dx$$ which represents the volume in one octant, where the $\alpha \beta$ term controls the orientation of the definite integrals. This evaluates to $1/6$ irrespective of the differing values $\alpha, \beta$ for an octant, so $ V=8 \cdot (1/6) = 4/3 $.
For the surface area consider $$\alpha \beta \int_0^{\alpha} \int_0^{\frac{(1-\alpha x)}{\beta}} \sqrt{1+\frac{\alpha^2}{\gamma^2}+\frac{\beta^2}{\gamma^2}}dy dx$$ $$=\alpha \beta \int_0^{\alpha} \int_0^{\frac{(1-\alpha x)}{\beta}} \sqrt{3}dy dx$$ which comes from the usual surface area integal with $z=\frac{1-\alpha x - \beta y}{\gamma}$. This evaluates to $\frac{\sqrt{3}}{2} $ for every octant, giving $SA = 4\sqrt{3}$.
From the equation, you can find the corners of your solid to be $(\pm 1, 0, 0)$, $(0, \pm 1, 0)$, $(0, 0, \pm 1)$. Since the equation is linear, you can imagine your solid as the interior of planes connecting these six corners. Using a bit of imagination, it becomes clear that the solid in question is an octahedron with edge length $\sqrt{2}$. The area is $\frac{\sqrt{3}}{4}\cdot(\sqrt{2})^2\cdot8=4 \sqrt{3}$ ($8$ equilateral triangles).
As for the volume, you can find the volume in a single quadrant and multiply by $8$. Volume in a quadrant: $\int_{0}^{1} \frac{1}{2} x^2 dx = \frac{1}{6}$. Total volume: $\frac{1}{6} \cdot 8 = \frac{4}{3}$.