Question

A light bulb is designed by revolving the graph of:

$y = \frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}$

Over the interval 0 ≤ x ≤ 1/3 about the x-axis, where x and y are measured in feet. Find the surface area of the bulb and use the result to approximate the amount of glass needed to make the bulb. (Glass is 0.015 inch thick).​

Answer 1

The graph revolved is,

$\longrightarrow y=\dfrac{1}{3}\,x^{\frac{1}{2}}-x^{\frac{3}{2}}$

Consider an element on the graph which is,

• at a distance of x feet from y axis.

• at a distance of y feet from x axis.

• ds feet wide.

The width ds is measured along the graph and is given by,

$\longrightarrow ds=\sqrt{(dx)^2+(dy)^2}$

because dx, dy and ds form a right triangle. In other words ds is contributed by the infinitesimal change on the graph horizontally and vertically.

We can rewrite ds as,

$\longrightarrow ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\ dx$

Let us find dy/dx.

$\longrightarrow y=\dfrac{1}{3}\,x^{\frac{1}{2}}-x^{\frac{3}{2}}$

$\longrightarrow\dfrac{dy}{dx}=\dfrac{1}{6}\,x^{-\frac{1}{2}}-\dfrac{3}{2}x^{\frac{1}{2}}$

Then,

$\longrightarrow ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\ dx$

$\longrightarrow ds=\sqrt{1+\left(\dfrac{1}{6}\,x^{-\frac{1}{2}}-\dfrac{3}{2}\,x^{\frac{1}{2}}\right)^2}\ dx$

$\longrightarrow ds=\sqrt{1-\dfrac{1}{2}+\dfrac{1}{36}\,x^{-1}+\dfrac{9}{4}\,x}\ dx$

$\longrightarrow ds=\sqrt{\dfrac{1}{2}+\dfrac{1}{36}\,x^{-1}+\dfrac{9}{4}\,x}\ dx$

$\longrightarrow ds=\left(\dfrac{1}{6}\,x^{-\frac{1}{2}}+\dfrac{3}{2}\,x^{\frac{1}{2}}\right)\ dx$

As the graph is revolved about x axis, the element considered makes a circle of radius y feet about the x axis.

So the surface area of the element will be,

$\longrightarrow dA=2\pi y\ ds$

$\longrightarrow dA=2\pi\left(\dfrac{1}{3}\,x^{\frac{1}{2}}-x^{\frac{3}{2}}\right)\left(\dfrac{1}{6}\,x^{-\frac{1}{2}}+\dfrac{3}{2}\,x^{\frac{1}{2}}\right)\ dx$

$\longrightarrow dA=2\pi\left(\dfrac{1}{18}+\dfrac{1}{3}\,x-\dfrac{3}{2}\,x^2\right)\ dx$

$\longrightarrow dA=\left(\dfrac{\pi}{9}+\dfrac{2\pi}{3}\,x-3\pi x^2\right)\ dx$

Then the total surface of the bulb is given by,

$\displaystyle\longrightarrow A=\int\limits_0^{\frac{1}{3}}\left(\dfrac{\pi}{9}+\dfrac{2\pi}{3}\,x-3\pi x^2\right)\ dx$

$\displaystyle\longrightarrow A=\dfrac{\pi}{9}\big[x\big]_0^{\frac{1}{3}}+\dfrac{\pi}{3}\left[x^2\right]_0^{\frac{1}{3}}-\pi\left[x^3\right]_0^{\frac{1}{3}}$

$\displaystyle\longrightarrow A=\dfrac{\pi}{9}\cdot\dfrac{1}{3}+\dfrac{\pi}{3}\cdot\dfrac{1}{9}-\pi\cdot\dfrac{1}{27}$

$\displaystyle\longrightarrow A=\dfrac{\pi}{27}+\dfrac{\pi}{27}-\dfrac{\pi}{27}$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{\pi}{27}\ \rm{feet}^2}}}}$

$\displaystyle\longrightarrow\underline{\underline{A=0.116355\ \rm{feet^2}}}}}$

$\displaystyle\longrightarrow A=0.116355\times144\ \rm{inch^2}}}$

$\displaystyle\longrightarrow\underline{\underline{A=16.755161\ \rm{inch^2}}}}}$

To approximate the amount of glass needed to make the bulb, we need to compute the volume of the glass required, which is equal to the volume of the outer cover of the bulb. The area of the glass required is the same as A.

Thickness of glass, d = 0.015 inch.

So,

$\longrightarrow V=Ad$

$\displaystyle\longrightarrow V=16.755161\times0.015\ \rm{inch^3}}}$

$\displaystyle\longrightarrow\underline{\underline{V=0.251327\ \rm{inch^3}}}}}$