Q.46 Each of the integrals or integra... [FREE SOLUTION]

Chapter 13: Q.46 (page 991)

Each of the integrals or integral expressions in Exercises 39鈥�46 represents the volume of a solid in ${鈩�}^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions
$2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 - 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃$

Short Answer

Expert verified

The value of integral is
$2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 - 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 = \frac{3}{2}$

Step by step solution

Given information

The given expressions

$2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 - 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 = \frac{3}{2}$

Simplificaiton

Here, $r = 0, r = (\sqrt{2} / 2) + \sin 胃$ and $胃 = - 蟺 / 4, 胃 = - 蟺 / 2, 胃 = 蟺 / 2$

Suppose

$I = 2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 - 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃$

$I = I_{1} - I_{2}$

Where,

$I_{1} = 2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {鈭�}_{0}^{(\sqrt{2} / 2) + \sin 胃} r d r d 胃 I_{1} = 2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {[\frac{r^{2}}{2}]}_{0}^{(\sqrt{2} / 2) + \sin 胃} d 胃 I_{1} = 2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} [\frac{{(\sqrt{2} / 2) + \sin 胃}^{2} - 0}{2}] 胃$

Now

$I_{2} = 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} {鈭�}_{0}^{(\sqrt{2} / 2) r \sin 胃} r d r d 胃 I_{2} = 2 {鈭�}_{- 蟺 / 2}^{- x / 4} {[\frac{r^{2}}{2}]}_{0}^{(\sqrt{2} / 2) + \sin 胃} d 胃 I_{2} = 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} [\frac{\{\frac{1}{2} + \sqrt{2} \sin 胃 + \frac{1}{2} (1 - \cos 2 胃)\}}{2}] 胃$

$I_{2} = 2 {[\frac{\{胃 - \sqrt{2} \cos 胃 - \frac{1}{4} \sin 2 胃\}}{2}]}_{- 蟺 / 2}_{}^{- 蟺 / 4} = 2 [\frac{\{- \frac{蟺}{4} - \sqrt{2} \cos (- \frac{蟺}{4}) - \frac{1}{4} \sin (- \frac{蟺}{2})\} - \{- \frac{蟺}{2} - \sqrt{2} \cos (- \frac{蟺}{2}) - \frac{1}{4} \sin (- 蟺)\}}{2}]$

$I_{2} = 2 [\frac{\{- \frac{蟺}{4} - 1 + \frac{1}{4}\} - \{- \frac{蟺}{2}\}}{2}]$

$I_{2} = \frac{3}{4} (蟺 - 1)$

Therefore,

$I = \frac{3}{4} (蟺 + 1) - \frac{3}{4} (蟺 - 1) I = \frac{3}{2}$

Thus, the value of integral is

$2 {鈭�}_{- 蟺 / 4}^{蟺 / 2} {鈭�}_{0}^{(\sqrt{2} / 2) 路 \sin 胃} r d r d 胃 - 2 {鈭�}_{- 蟺 / 2}^{- 蟺 / 4} {鈭�}_{0}^{(\sqrt{2} / 2) r \sin 胃} r d r d 胃 = \frac{3}{2}$

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Each of the integrals or integral expressions in Exercises 39鈥�46 represents the volume of a solid in 鈩�3. Use polar coordinates to describe the solid, and evaluate the expressions2鈭�-蟺/4蟺/2鈭�0(2/2)+sin胃rdrd胃-2鈭�-蟺/2-蟺/4鈭�0(2/2)+sin胃rdrd胃

Short Answer

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