Chapter 13: Q.46 (page 991)

Each of the integrals or integral expressions in Exercises 39–46 represents the volume of a solid in $ℝ^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions
$2 \int_{- π / 4}^{π / 2} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ - 2 \int_{- π / 2}^{- π / 4} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ$

Short Answer

Expert verified

The value of integral is
$2 \int_{- π / 4}^{π / 2} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ - 2 \int_{- π / 2}^{- π / 4} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ = \frac{3}{2}$

Step by step solution

Given information

The given expressions

$2 \int_{- π / 4}^{π / 2} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ - 2 \int_{- π / 2}^{- π / 4} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ = \frac{3}{2}$

Simplificaiton

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

Suppose

$I = 2 \int_{- π / 4}^{π / 2} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ - 2 \int_{- π / 2}^{- π / 4} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ$

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

Where,

$I_{1} = 2 \int_{- π / 4}^{π / 2} \int_{0}^{(\sqrt{2} / 2) + \sin θ} r d r d θ I_{1} = 2 \int_{- π / 4}^{π / 2} {[\frac{r^{2}}{2}]}_{0}^{(\sqrt{2} / 2) + \sin θ} d θ I_{1} = 2 \int_{- π / 4}^{π / 2} [\frac{{(\sqrt{2} / 2) + \sin θ}^{2} - 0}{2}] θ$

Now

$I_{2} = 2 \int_{- π / 2}^{- π / 4} \int_{0}^{(\sqrt{2} / 2) r \sin θ} r d r d θ I_{2} = 2 \int_{- π / 2}^{- x / 4} {[\frac{r^{2}}{2}]}_{0}^{(\sqrt{2} / 2) + \sin θ} d θ I_{2} = 2 \int_{- π / 2}^{- π / 4} [\frac{\{\frac{1}{2} + \sqrt{2} \sin θ + \frac{1}{2} (1 - \cos 2 θ)\}}{2}] θ$

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

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

$I_{2} = \frac{3}{4} (π - 1)$

Therefore,

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

Thus, the value of integral is

$2 \int_{- π / 4}^{π / 2} \int_{0}^{(\sqrt{2} / 2) \cdot \sin θ} r d r d θ - 2 \int_{- π / 2}^{- π / 4} \int_{0}^{(\sqrt{2} / 2) r \sin θ} r d r d θ = \frac{3}{2}$

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Each of the integrals or integral expressions in Exercises 39–46 represents the volume of a solid in ℝ3. Use polar coordinates to describe the solid, and evaluate the expressions2∫-π/4π/2∫0(2/2)+sinθrdrdθ-2∫-π/2-π/4∫0(2/2)+sinθrdrdθ

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