Chapter 6: Q. 83 (page 541)

Use Theorem 6.10 to prove that a sphere of radius $r$ has surface area $4 {πr}^{2}$ .

Short Answer

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Sphere of radius $r$ has surface area $4 {πr}^{2}$ .

Step by step solution

Step 1. Given Information

Sphere of radius $r$ .

Step 2. Finding the surface area of sphere

$The surface area of the solid of revolution obtained by revolving f (x) around the x - axis from x = a to x = b is : 2 π \int_{a}^{b} f (x) \sqrt{1 + (f' {(x))}^{2}} d x Equation of circle of radius r i s : x^{2} + y^{2} = r^{2} Therefore, y = \sqrt{r^{2} - x^{2}} = f (x) and f' (x) = - \frac{x}{\sqrt{r^{2} - x^{2}}} Surface area of sphere is surface area of the solid of revolution obtained by revolving f (x) around the x - axis from x = - r to x = r Therefore, S = 2 π \int_{- r}^{r} f (x) \sqrt{1 + (f' {(x))}^{2}} d x C = 2 π \int_{- r}^{r} \sqrt{r^{2} - x^{2}} \sqrt{1 + {(- \frac{x}{\sqrt{r^{2} - x^{2}}})}^{2}} d x C = 2 π \int_{- r}^{r} \sqrt{r^{2} - x^{2}} \sqrt{\frac{r^{2} - x^{2} + x^{2}}{r^{2} - x^{2}}} d x C = 2 π \int_{- r}^{r} r d x C = 2 πr [x_{- r}^{r}] C = 2 πr [r - (- r)] C = 2 πr [2 r] C = 4 {πr}^{2}$

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Use Theorem 6.10 to prove that a sphere of radius $r$ has surface area $4 {πr}^{2}$ .

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the surface area of sphere

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Use Theorem 6.10 to prove that a sphere of radius r has surface area4πr2.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the surface area of sphere

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Applied Mathematics

Geometry

Discrete Mathematics

Statistics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Use Theorem 6.10 to prove that a sphere of radius $r$ has surface area $4 {πr}^{2}$ .