Q 15. The following integral expressio... [FREE SOLUTION]

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Chapter 9: Q 15. (page 756)

The following integral expression may be used to find the area of a region in the polar coordinate plane: $\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃$
Sketch the region and then compute its area. (If you prefer, you may use a simpler integral to compute the same area.)

Short Answer

Expert verified

$(\frac{蟺}{8} - \frac{1}{4})$

Step by step solution

Given information

$\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃$

Calculation

Consider the integral, $\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃$

The goal is to determine the integral's value.

Take the integral, $\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃$

Then,

$\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃 = \frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \frac{1 - \cos 2 胃}{2} d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \frac{1 + \cos 2 胃}{2} d 胃 [since \cos^{2} 胃 = \frac{1 + \cos 2 胃}{2}, \sin^{2} 胃 = \frac{1 - \cos 2 胃}{2}] \frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃 = \frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \frac{1}{2} - \frac{\cos 2 胃}{2} d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \frac{1}{2} + \frac{\cos 2 胃}{2} d 胃$

[since \cos^{2} 胃 = \frac{1 + \cos 2 胃}{2}, \sin^{2} 胃 = \frac{1 - \cos 2 胃}{2}] \frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃 = \frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \frac{1}{2} - \frac{\cos 2 胃}{2} d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \frac{1}{2} + \frac{\cos 2 胃}{2} d 胃 = \frac{1}{2} {[\frac{胃}{2} - \frac{\sin 2 胃}{2 路 2}]}_{0}^{\frac{蟺}{4}} + \frac{1}{2} {[\frac{胃}{2} + \frac{\sin 2 胃}{2 路 2}]}_{\frac{蟺}{4}}^{\frac{蟺}{2}}

By applying the limits,

\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃 = \frac{1}{2} [\frac{蟺}{8} - \frac{\sin 2 路 \frac{蟺}{4}}{2 路 2} - 0 - \frac{\sin 2 路 0}{2 路 2}] + \frac{1}{2} [\frac{蟺}{4} + \frac{\sin 2 路 \frac{蟺}{2}}{2 路 2} - \frac{蟺}{8} - \frac{\sin 2 路 \frac{蟺}{4}}{2 路 2}] = \frac{1}{2} [\frac{蟺}{8} - \frac{1}{4} - 0 - 0] + \frac{1}{2} [\frac{蟺}{4} + 0 - \frac{蟺}{8} - \frac{1}{4}] = \frac{1}{2} [\frac{蟺}{8} - \frac{1}{4} + \frac{蟺}{4} - \frac{蟺}{8} - \frac{1}{4}]

Calculation

Thus,

$\frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃 = \frac{1}{2} [\frac{蟺}{4} - \frac{2}{4}] \frac{1}{2} {鈭�}_{0}^{\frac{蟺}{4}} \sin^{2} 胃 d 胃 + \frac{1}{2} {鈭�}_{\frac{蟺}{4}}^{\frac{蟺}{2}} \cos^{2} 胃 d 胃 = [\frac{蟺}{8} - \frac{1}{4}]$

Therefore, the value of the integral is $(\frac{蟺}{8} - \frac{1}{4})$

Calculation

The graphical representation is as follows.

This is the graphical representation.

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The following integral expression may be used to find the area of a region in the polar coordinate plane: 12鈭�0蟺4sin2胃d胃+12鈭�蟺4蟺2cos2胃d胃Sketch the region and then compute its area. (If you prefer, you may use a simpler integral to compute the same area.)

Short Answer

Step by step solution

Given information

Calculation

Calculation

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Geometry

Discrete Mathematics

Mechanics Maths

Decision Maths

Pure Maths

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