Volume of all points within $K : x^2 + y^2 + z^2 \leq 1$ with a maximum distance of $\sqrt{2}$ to the point P on the boundary of K.

Question

Let $K : x^2 + y^2 + z^2 \leq 1$ and $P = (a,b,c)$ be a point on the border of of K (i.e $a^2 + b^2 + c^2 = 1)$.

Determine the volume of the body of all points within $K$ with a maximum distance of $\sqrt{2}$ to the point P.

I know I can solve this by using a triple integral to calculate the volume between the sphere centered at $(0,0,1)$ with radius $\sqrt{2}$ and the sphere $x^2 + y^2 + z^2 \leq 1$.

However, I initially used a different method and I'm wondering why it's wrong. Here is my attempt:

The point $(x,y,z)$ must satisfy: $$x^2 + y^2 + z^2 \leq 1$$ $$(x-a)^2 + (y-b)^2 + (z-c)^2 \leq 2$$ The last condition is equivalent to: $$a^2 + b^2 + c^2 -2ax - 2by - 2cz + x^2 + y^2 + z^2 \leq 2$$ Since $a^2 + b^2 + c^2 = 1$, we get: $$x^2 + y^2 + z^2 -2(ax+by+cz) \leq 1 $$ Now, the tangent plane to the sphere in point P is equivalent to $ax+by+cz = C$ where $C$ is a constant. Since $(a,b,c)$ must satisfy the equation of the tangent plane we get that $C=1$. Substituting this into the above equation I get: $$x^2 + y^2 + z^2 \leq 3$$ This is obviously wrong since this volume is larger than the initial sphere. What have I actually calculated? Why is my method wrong?

Answer 1

In your derivation, you assume the existence of a particular point on the surface of the smaller sphere satisfying $a^2 + b^2 + c^2 = 1$. You then use the following facts: $$ (x-a)^2 + (y-b)^2 + (z-c)^2 \leq 2 \\ ax + by + cz = 1 $$ You are then describing all points that satisfy both of these conditions. This is the intersection of a ball with a plane with a circle of radius $\sqrt{2}$; in other words, it's a flat circular disc of radius $\sqrt{2}$ centered at $(a,b,c)$ (and orthogonal to this vector.)

Once you see this, it's not too hard to see that nothing is "wrong" with your result $x^2 + y^2 + z^2 \leq 3$, since all points on this disc satisfy this inequality. It's just not the set you're looking for.

EDIT: Here's a rendering of what's going on, with $(a,b,c) = (1,0,0)$. Your inequality is true for the shaded disc on the right.

Answer 2

Your mistake is the following: You arrived at the inequality $$x^2+y^2+z^2-2(ax+by+cz)\leq1$$ that any point $(x,y,z)$ of the solid $B$ must satisfy. Now you suddenly use $(x,y,z)$ as variable for points in the tangent plane $T$ at $P$ and say that all points in $T$ have to satisfy $ax+by+cz=1$ (which is true). But with the exception of $P$ no point of $B$ lies in $T$, hence the condition $ax+by+cz=1$ is totally irrelevant for the problem.

By the way: You don't need calculus for this problem. The solid $B$ in question is the union of half a ball of radius $1$ and a cap of height $h=\sqrt{2}-1$ from a ball of radius $\sqrt{2}$. There are elementary formulas for the corresponding volumina.