Chapter 14: Q. 46 (page 1132)

Evaluate the integrals in Exercises 43–46 directly or using Green’s Theorem.
$\iint_{R} (3 x y - 4 x^{2} y) d A$ , where R is the unit disk.

Short Answer

Expert verified

Ans: The required integral is $\iint_{R} (3 x y - 4 x^{2} y) d A = 0$ .

Step by step solution

Step 1. Given information:

$\iint_{R} (3 x y - 4 x^{2} y) d A$

Step 2. Finding the region of integration

Here, the region R is the unit disk. In polar coordinates, the region of integration is described as follows,

$R = {(r, θ) ∣ 0 \leq r \leq 1, 0 \leq θ \leq 2 π} .$

In this case,

$x = r \cos θ, y = r \sin θ, x^{2} + y^{2} = r^{2}, and d A = r d r d θ .$

Step 3. Evaluating the integral:

Then, evaluate the required integral as follows:

$\iint_{R} (3 x y - 4 x^{2} y) d A = \int_{0}^{2 π} \int_{0}^{1} (3 (r \cos θ) (r \sin θ) - 4 (r \cos θ)^{2} (r \sin θ)) r d r d θ = \int_{0}^{2 π} \int_{0}^{1} (3 r^{2} \sin θ \cos θ - 4 r^{3} \sin θ \cos^{2} θ) r d r d θ = \int_{0}^{2 π} \int_{0}^{1} (3 r^{3} \sin θ \cos θ - 4 r^{4} \sin θ \cos^{2} θ) d r d θ = \int_{0}^{2 π} [\int_{0}^{1} (3 r^{3} \sin θ \cos θ - 4 r^{4} \sin θ \cos^{2} θ) d r] d θ = \int_{0}^{2 π} {[\frac{3 r^{4}}{4} \sin θ \cos θ - \frac{4 r^{5}}{5} \sin θ \cos^{2} θ]}_{0}^{1} d θ = \int_{0}^{2 π} [(\frac{3 \cdot 1^{4}}{4} \sin θ \cos θ - \frac{4 \cdot 1^{5}}{5} \sin θ \cos^{2} θ) - (\frac{3 \cdot 0^{4}}{4} \sin θ \cos θ - \frac{4 \cdot 0^{5}}{5} \sin θ \cos^{2} θ)] d θ = \int_{0}^{2 π} [(\frac{3}{4} \sin θ \cos θ - \frac{4}{5} \sin θ \cos^{2} θ) - 0] d θ = \int_{0}^{2 π} (\frac{3}{4} \cos θ - \frac{4}{5} \cos^{2} θ) \sin θ d θ = - {[\frac{3}{4} \cdot \frac{\cos^{2} θ}{2} - \frac{4}{5} \cdot \frac{\cos^{3} θ}{3}]}_{0}^{2 π} = - [(\frac{3}{4} \cdot \frac{\cos^{2} 2 π}{2} - \frac{4}{5} \cdot \frac{\cos^{3} 2 π}{3}) - (\frac{3}{4} \cdot \frac{\cos^{2} 0}{2} - \frac{4}{5} \cdot \frac{\cos^{3} 0}{3})] = - [(\frac{3}{4} \cdot \frac{1^{2}}{2} - \frac{4}{5} \cdot \frac{1^{3}}{3}) - (\frac{3}{4} \cdot \frac{1^{2}}{2} - \frac{4}{5} \cdot \frac{1^{3}}{3})] = 0$

Therefore, the required integral is $\iint_{R} (3 x y - 4 x^{2} y) d A = 0$ .

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Evaluate the integrals in Exercises 43–46 directly or using Green’s Theorem.
$\iint_{R} (3 x y - 4 x^{2} y) d A$ , where R is the unit disk.

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Finding the region of integration

Step 3. Evaluating the integral:

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Evaluate the integrals in Exercises 43–46 directly or using Green’s Theorem.∬R3xy-4x2ydA, where R is the unit disk.

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Finding the region of integration

Step 3. Evaluating the integral:

One App. One Place for Learning.

Most popular questions from this chapter

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Statistics

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Study anywhere. Anytime. Across all devices.

Evaluate the integrals in Exercises 43–46 directly or using Green’s Theorem.
$\iint_{R} (3 x y - 4 x^{2} y) d A$ , where R is the unit disk.