Q. 83 Use Theorem 6.10 to prove that a... [FREE SOLUTION]

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Chapter 6: Q. 83 (page 541)

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

Short Answer

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

Step by step solution

Step 1. Given Information

Sphere of radius $r$ .

Step 2. Finding the surface area of sphere

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

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91影视

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

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the surface area of sphere

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91影视

Use Theorem 6.10 to prove that a sphere of radius r has surface area4蟺谤2.

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

Pure Maths

Probability and Statistics

Geometry

Applied Mathematics

Statistics

Study anywhere. Anytime. Across all devices.

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