Q. 3 Each of the integral expressions... [FREE SOLUTION]

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Chapter 13: Q. 3 (page 1025)

Each of the integral expressions that follow represents the area of a region in the plane bounded by a function expressed in polar coordinates. Use the ideas from this section and from Chapter 9 to sketch the regions, and then evaluate each integral
${鈭�}_{0}^{\frac{2 蟺}{3}} {(\frac{1}{2} + \cos 胃)}^{2} d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} {(\frac{1}{2} + \cos 胃)}^{2} d 胃$

Short Answer

Expert verified

The value of integral is $\frac{蟺}{4}$

Step by step solution

.Given information

Integral;

${鈭�}_{0}^{\frac{2 蟺}{3}} {(\frac{1}{2} + \cos 胃)}^{2} d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} {(\frac{1}{2} + \cos 胃)}^{2} d 胃$

Step 2. Plot the region:

In the given integral we can see that there is a subtraction occurs between integral.

By comparing the given integral with the formula of area of region of the polar curve $r = f (胃)$ :

$\frac{1}{2} {鈭�}_{a}^{b} r^{2} d 胃$

We get $r = \sqrt{2} (\frac{1}{2} + \cos 胃)$

Now plot this curve in both given interval of $胃$

Step 3. Simplify integral

${鈭�}_{0}^{\frac{2 蟺}{3}} {(\frac{1}{2} + \cos 胃)}^{2} d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} {(\frac{1}{2} + \cos 胃)}^{2} d 胃 = {鈭�}_{0}^{\frac{2 蟺}{3}} (\frac{1}{4} + \cos^{2} 胃 + \cos 胃) d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} (\frac{1}{4} + \cos^{2} 胃 + \cos 胃) d 胃 = {鈭�}_{0}^{\frac{2 蟺}{3}} (\frac{1}{4} + \frac{1 + \cos 2 胃}{2} + \cos 胃) d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} (\frac{1}{4} + \frac{1 + \cos 2 胃}{2} + \cos 胃) d 胃 = {鈭�}_{0}^{\frac{2 蟺}{3}} (\frac{1}{4} + \frac{1}{2} + \frac{\cos 2 胃}{2} + \cos 胃) d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} (\frac{1}{4} + \frac{1}{2} + \frac{\cos 2 胃}{2} + \cos 胃) d 胃 = {鈭�}_{0}^{\frac{2 蟺}{3}} (\frac{3}{4} + \frac{\cos 2 胃}{2} + \cos 胃) d 胃 - {鈭�}_{蟺}^{\frac{4 蟺}{3}} (\frac{3}{4} + \frac{\cos 2 胃}{2} + \cos 胃) d 胃$

Step 4. Solve integral

Now integrall can be solved as

{[\frac{3}{4} 胃 + \frac{1}{2} \frac{s i n 2 胃}{2} + \sin 胃]}_{0}^{\frac{2 蟺}{3}} - {[\frac{3}{4} 胃 + \frac{1}{2} \frac{s i n 2 胃}{2} + \sin 胃]}_{蟺}^{\frac{4 蟺}{3}} = [\frac{3 脳 \frac{2 蟺}{3}}{4} - \frac{1}{4} \sin (\frac{4 蟺}{3}) + \sin (\frac{2 蟺}{3}) - 0] - [\frac{3 脳 \frac{4 蟺}{3}}{4} - \frac{1}{4} \sin (\frac{8 蟺}{3}) + \sin (\frac{4 蟺}{3}) - (\frac{3}{4} 蟺 + \frac{1}{4} \sin (2 蟺) + 蝉颈苍蟺)] = (\frac{蟺}{2} - \frac{\sqrt{3}}{8} + \frac{\sqrt{3}}{2}) - (蟺 - \frac{\sqrt{3}}{8} + \frac{\sqrt{3}}{2} - \frac{3}{4} 蟺 - 0 - 0) = \frac{蟺}{2} - \frac{\sqrt{3}}{8} + \frac{\sqrt{3}}{2} - 蟺 + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{2} + \frac{3}{4} 蟺 = \frac{3 蟺 - 2 蟺}{4} = \frac{蟺}{4}

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Short Answer

Step by step solution

.Given information

Step 2. Plot the region:

Step 3. Simplify integral

Step 4. Solve integral

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