Mr. Mendrin I presume

Calculus Level 3

The below object's surface area and the volume defined by the region lying inside the cylinder $x^2 + y^2 = 1$ and inside the sphere $(x - 1)^2 + y^2 + z^2 = 4$ can be represented by

$\pi \cdot A^3 \quad \text{ and } \quad \pi \cdot \dfrac{A^4}B - \dfrac{A^6}{B^2} ,$

respectively, where $A$ and $B$ are coprime positive integers.

Find $A+B$ .

The answer is 5.

1 solution

Michael Mendrin
Nov 8, 2016

First, shift the equations so that the sphere is centered at $\left(0,0,0\right)$ . Then we integrate by slices parallel to the $xz$ plane. The radius of the sphere sliced by a plane at distance $x$ from the origin is

$r=\sqrt { { 2 }^{ 2 }-{ x }^{ 2 } }$

Meanwhile, the $y$ of the unit circle forming the cylinder is

$y=\sqrt { { 1 }^{ 2 }-{ (x-1) }^{ 2 } }$

So that we integrate the areas of the slices as follows

$\displaystyle \int _{ 0 }^{ 2 }{ 2\left( { r }^{ 2 }ArcSin\left( \dfrac { y }{ r } \right) +y\sqrt { { r }^{ 2 }-{ y }^{ 2 } } \right) dx } =\dfrac { 16 }{ 3 } \pi -\dfrac { 64 }{ 9 }$

And so $a=2$ and $b=3$ , and the answer is $5$

Mr. Mendrin I presume

The answer is 5.

1 solution

0 pending reports