Q. 82 - Calculus

Question

Use Theorem 6.10 to prove that a sphere of radius 5 has surface area $100\mathrm{\pi }$.

Step-by-Step Solution

Answer

Sphere of radius 5 has surface area $100\mathrm{\pi }$.

Step 1. Given Information

Sphere of radius 5.

Step 2. Finding the surface area of sphere

$$\begin{gathered} \mathrm{The} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{solid} \mathrm{of} \mathrm{revolution} \mathrm{obtained} \mathrm{by} \mathrm{revolving} f\left(x\right) \\ \mathrm{around} \mathrm{the} x-\mathrm{axis} \mathrm{from} x=a \mathrm{to} x=b \mathrm{is}: 2\pi {\int }_a^{b}f\left(x\right) \sqrt{1 + ( f\text{'}{\left(x\right))}^2}dx \\ \mathrm{Equation} \mathrm{of} \mathrm{circle} \mathrm{of} \mathrm{radius} r is: x^2+y^2=5^2 \\ \mathrm{Therefore}, y=\sqrt{5^2-x^2}=f\left(x\right) \\ \mathrm{and} f\text{'}\left(x\right)=-\frac{x}{\sqrt{5^2-x^2}} \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{sphere} \mathrm{is} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{solid} \mathrm{of} \mathrm{revolution} \mathrm{obtained} \\ \mathrm{by} \mathrm{revolving} \mathrm{f}\left(\mathrm{x}\right) \mathrm{around} \mathrm{the} \mathrm{x}-\mathrm{axis} \mathrm{from} \mathrm{x}=-5 \mathrm{to} \mathrm{x}=5 \\ \mathrm{Therefore}, S=2\mathrm{\pi }{\int }_{-5}^5\mathrm{f}\left(\mathrm{x}\right)\sqrt{1 + ( \mathrm{f}\text{'}{\left(\mathrm{x}\right))}^2}d\mathrm{x} \\ \mathrm{C}=2\mathrm{\pi }{\int }_{-5}^5\sqrt{5^2-{\mathrm{x}}^2}\sqrt{1 + {\left(-\frac{\mathrm{x}}{\sqrt{5^2-{\mathrm{x}}^2}}\right)}^2}d\mathrm{x} \\ \mathrm{C}=2\mathrm{\pi }{\int }_{-5}^5\sqrt{5^2-{\mathrm{x}}^2}\sqrt{\frac{5^2-{\mathrm{x}}^2+{\mathrm{x}}^2}{5^2-{\mathrm{x}}^2}}d\mathrm{x} \\ \mathrm{C}=2\mathrm{\pi }{\int }_{-5}^{5}5 d\mathrm{x} \\ \mathrm{C}=10\pi \left[{\overline{)\mathrm{x}}}_{-5}^5\right] \\ \mathrm{C}=10\pi \left[5-(-5)\right] \\ \mathrm{C}=10\pi \left[10\right] \\ \mathrm{C}=100\pi \end{gathered}$$