volume of solid whose distance is at most t, from a given parallelepiped

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volume of solid whose distance is at most t, from a given parallelepiped

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#1 h May 24, 2019, 11:03 AM • 3 Y

Y by kiyoras_2001, Adventure10, Mango247

Given a parallelepiped $P$ , let $V_P$ be its volume, $S_P$ the area of its surface and $L_P$ the sum of the lengths of its edges. For a real number $t \ge 0$ , let $P_t$ be the solid consisting of all points $X$ whose distance from some point of $P$ is at most $t$ . Prove that the volume of the solid $P_t$ is given by the formula $V(P_t) =V_P + S_Pt + \frac{\pi}{4} L_P t^2 + \frac{4\pi}{3} t^3$ .

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#2 h Jan 11, 2020, 1:44 PM • 1 Y

Y by Adventure10

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#3 h Jun 26, 2020, 4:58 PM

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The solid $P_t$ consists of
1. The parallelepiped $P$ with volume $V_P$ ;
2. Pieces of right prisms constructed on the faces of $P$ with heights $h$ that supply $S_Pt$ total volume;
3. Pieces of cylinders leaning to the edges of $P$ . Note that the four pieces of cylinders corresponding to some four parallel edges of $P$ form a whole cylinder. Summing up the volume of this cylinder with the volumes of the other two cylinders corresponding to the other two quartets of parallel edges we get $\frac{\pi}{4}L_Pt^2$ total volume;
4. Pieces of balls centered at the vertices of $P$ . Again note that the eight pieces of the balls together form a complete ball of radius $t$ whence the volumes of the ball pieces sum up to $\frac{4}{3}\pi t^3$ .
Adding the calculated volumes we get
$V(P_t)=V_P+S_Pt+\frac{\pi}{4}L_Pt^2+\frac{4}{3}\pi t^3$ as wanted.

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