Question

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

Short answer

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

Step-by-step solution

Step 1. Given Information  

Sphere of radius 5.

Step 2. Finding the surface area of sphere

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