Q. 62 - Calculus

Question

Mark crafts wooden spheres into napkin rings. His napkin rings are depicted next to the left with height $h=8$ and radius $R=5$ in millimeters. Alina also makes sphere-based napkin rings; however, because she prefers larger napkins, she makes them with the height $h=8$ and radius $R=6$ seen to the right. Show that Mark and Alina use the same quantity of wood to make their napkin rings by using the shell approach.

Step-by-Step Solution

Answer

Both the napkin rings used the same amount of wood. 

Step 1: Given information

The following schematics show the cross section of the two napkin rings.

Step 2: Calculation

By rotating the circular arc AB about the y-axis and multiplying the result by two while the cross-section is schematically oriented about the x-axis, the volume of the first napkin ring may be calculated.

By rotating the circular arc CD about the y-axis and multiplying the result by two while the cross-section is schematically oriented about the x-axis, the volume of the second napkin ring may be calculated.

Both rings are eight inches tall. Consequently, both of the circular parts in the accompanying image have a height of 4.

The circle with radius 5 that is centered at the origin includes the circular arc AB.

As a result, the equation for the arc AB is defined as,

$\begin{array}{rcl}{x}^2+y^2& =& 25\\ y& =& \sqrt{25-x^2}\end{array}$

Here, x varies from 3 to 5.

The volume of the first napkin ring is calculated as,

$\begin{array}{rcl}{V}_1& =& 2·2\pi {\int }_3^{5}x\sqrt{25-x^2}dx\\ & =& 4\pi {\int }_3^{5}x\sqrt{25-x^2}dx\end{array}$

Substitute $25-x^2=t$ then, $-2 x d x=d t.$

The limits are,

style="width:30%" $\begin{array}{rcl}t& =& 25-3^2\\ & =& 16\\ t& =& 25-5^2\\ & =& 0\end{array}$

Thus, the integral style="width:30%" $V_1=4\pi {\int }_3^{5}x\sqrt{25-x^2}dx$ becomes,

style="width:30%" $\begin{array}{rcl}{V}_1& =& -2\pi {\int }_{16}^0\sqrt{t}dt\\ & =& -\frac{4\pi }{3}{\left[t^{3/2}\right]}_{16}^0\\ & =& \frac{256\pi }{3}\end{array}$

Step 3: Further Calculation

The circular arc CD, on the other hand, is a portion of the circle of radius 6 and is centered at the origin.

Consequently, the arc CD's equation is given by,

$\begin{array}{rcl}{x}^2+y^2& =& 36\\ y& =& \sqrt{36-x^2}\end{array}$

To determine the coordinates of point C, locate the location where the circle $x^2+y^2=36$ and the line $y=4$ connect.

style="width:30%" $\begin{array}{rcl}{x}^2+4^2& =& 36\\ x& =& \sqrt{36-16}\\ x& =& 2\sqrt{5}\end{array}$

Therefore, x varies $2\sqrt{5}$ to $6$.

The volume of the second napkin ring is calculated as,

style="width:30%" $\begin{array}{rcl}{V}_2& =& 2·2\pi {\int }_{2\sqrt{5}}^{6}x\sqrt{36-x^2}dx\\ & =& 4\pi {\int }_{2\sqrt{5}}^{6}x\sqrt{36-x^2}dx\end{array}$

Substitute $36-x^2=t$ then, style="width:30%" $-2xdx=dt.$

The limits are,

style="width:30%" $\begin{array}{rcl}t& =& 36-{\left(2\sqrt{5}\right)}^2 \\ & =& 16\\ t& =& 36-6^2\\ & =& 0\end{array}$

Thus, the integral style="width:30%" $V_2=4\pi {\int }_{2\sqrt{5}}^{6}x\sqrt{36-x^2}dx$becomes

style="width:30%" $\begin{array}{rcl}{V}_2& =& -2\pi {\int }_{16}^0\sqrt{t}dt\\ & =& -\frac{4\pi }{3}{\left[t^{3/2}\right]}_{16}^0\\ & =& \frac{256\pi }{3}\end{array}$

The volume $V_1=\frac{256\pi }{3}$  and volume  $V_2=\frac{256\pi }{3}$ of both the napkin rings are equal.

Thus, both the napkin rings used the same amount of wood.