Evaluate the given integral by changing to polar coordinates: $\iint_R 7(x + y) \, dA$ where $R$ is the region that lies to the left of the $y$-axis between the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 16$.

Neetesh Kumar | November 28, 2024

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This is the solution to Math 1D
Assignment: 15.3 Question Number 2
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Step-by-step solution:

Step 1: Understanding the Problem

We are asked to evaluate the double integral over a region $R$, which is bounded by two circles:

• $x^2 + y^2 = 1$ (radius 1),
• $x^2 + y^2 = 16$ (radius 4).

The region $R$ lies to the left of the $y$-axis, meaning the bounds for $x$ are restricted to negative values (i.e., $x \leq 0$).

The function to integrate is $7(x + y)$, and we will convert this integral into polar coordinates.

Step 2: Converting to Polar Coordinates

In polar coordinates, the transformations are:

• $x = r \cos \theta$,
• $y = r \sin \theta$,
• The differential area element $dA = r \, dr \, d\theta$.

Also, the given function $7(x + y)$ becomes:

$7(x + y) = 7(r \cos \theta + r \sin \theta) = 7r(\cos \theta + \sin \theta)$

Step 3: Determining the Limits of Integration

The region $R$ is the area inside the annulus between the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 16$, and to the left of the $y$-axis. In polar coordinates:

• The radius $r$ ranges from $1$ to $4$ (since $r^2 = x^2 + y^2$ and the circles are defined by radii 1 and 4).
• The angle $\theta$ ranges from $\pi/2$ to $3\pi/2$ because we are only considering the left half of the plane, where $x \leq 0$.

Thus, the limits are:

• $r \in [1, 4]$,
• $\theta \in [\pi/2, 3\pi/2]$.

Step 4: Setting up the Integral in Polar Coordinates

Now that we have the transformations and limits, we can write the integral as:

$\iint_R 7(x + y) \, dA = \int_{\pi/2}^{3\pi/2} \int_1^4 7r (\cos \theta + \sin \theta) \, r \, dr \, d\theta$

This simplifies to:

$\int_{\pi/2}^{3\pi/2} \int_1^4 7r^2 (\cos \theta + \sin \theta) \, dr \, d\theta$

Step 5: Evaluating the Inner Integral

First, evaluate the inner integral with respect to $r$:

$\int_1^4 7r^2 (\cos \theta + \sin \theta) \, dr$

Since $\cos \theta + \sin \theta$ is independent of $r$, we can treat it as a constant during integration. The integral of $r^2$ is:

$\int r^2 \, dr = \frac{r^3}{3}$

Now, evaluate this from $r = 1$ to $r = 4$:

$\left[ \frac{r^3}{3} \right]_1^4 = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21$

Thus, the inner integral becomes:

$7 (\cos \theta + \sin \theta) \times 21 = 147 (\cos \theta + \sin \theta)$

Step 6: Evaluating the Outer Integral

Now, evaluate the outer integral with respect to $\theta$:

$\int_{\pi/2}^{3\pi/2} 147 (\cos \theta + \sin \theta) \, d\theta$

We can break this up into two integrals:

$147 \int_{\pi/2}^{3\pi/2} \cos \theta \, d\theta + 147 \int_{\pi/2}^{3\pi/2} \sin \theta \, d\theta$

1. Evaluating $\int_{\pi/2}^{3\pi/2} \cos \theta \, d\theta$:

$\int \cos \theta \, d\theta = \sin \theta$

Evaluating this from $\pi/2$ to $3\pi/2$:

$\sin(3\pi/2) - \sin(\pi/2) = -1 - 1 = -2$

2. Evaluating $\int_{\pi/2}^{3\pi/2} \sin \theta \, d\theta$:

$\int \sin \theta \, d\theta = -\cos \theta$

Evaluating this from $\pi/2$ to $3\pi/2$:

$-\cos(3\pi/2) + \cos(\pi/2) = 0 + 0 = 0$

Thus, the total integral becomes:

$147 \times (-2) + 147 \times 0 = -294$

Final Answer:

The value of the integral is: $\boxed{-294}$

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