Evaluate the given integral by changing to polar coordinates. $\iint_D x \, dA,$ where $D$ is the region in the first quadrant that lies between the circles $x^2 + y^2 = 4$ and $x^2 + y^2 = 2x$.

Neetesh Kumar | November 28, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 15.3 Question Number 6
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Step-by-step solution:

Step 1: Analyze the Region

The region $D$ is bounded by two circles:

1. The first circle is $x^2 + y^2 = 4$, which represents a circle with radius $2$ centered at the origin $(0, 0)$.
2. The second circle is $x^2 + y^2 = 2x$. To rewrite this in a more familiar form, we complete the square: $x^2 - 2x + y^2 = 0 \quad \text{or} \quad (x - 1)^2 + y^2 = 1.$ This represents a circle with radius $1$ centered at $(1, 0)$.

The region $D$ is the area in the first quadrant between these two circles. Therefore, we need to convert these equations and the limits of integration into polar coordinates.

Step 2: Convert to Polar Coordinates

In polar coordinates, we use the following transformations: $x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta.$

Now we convert the equations of the circles into polar form:

• For the first circle, $x^2 + y^2 = 4$, this becomes: $r^2 = 4 \quad \Rightarrow \quad r = 2.$

• For the second circle, $x^2 + y^2 = 2x$, this becomes: $r^2 = 2r \cos \theta \quad \Rightarrow \quad r = 2 \cos \theta.$ This represents the second circle in polar coordinates.

Step 3: Set up the Integral

The region $D$ is bounded by $r = 2 \cos \theta$ (the second circle) and $r = 2$ (the first circle). We also restrict ourselves to the first quadrant, so $\theta$ ranges from $0$ to $\frac{\pi}{2}$.

The integral we need to evaluate is: $\iint_D x \, dA = \iint_D r \cos \theta \, r \, dr \, d\theta = \int_0^{\frac{\pi}{2}} \int_{2\cos \theta}^{2} r^2 \cos \theta \, dr \, d\theta.$

Step 4: Evaluate the Inner Integral

We first evaluate the inner integral with respect to $r$: $\int_{2 \cos \theta}^{2} r^2 \, dr.$

The antiderivative of $r^2$ is $\frac{r^3}{3}$, so: $\int_{2 \cos \theta}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{2 \cos \theta}^{2} = \frac{8}{3} - \frac{8 \cos^3 \theta}{3}.$

Thus, the integral becomes: $\int_0^{\frac{\pi}{2}} \left( \frac{8}{3} - \frac{8 \cos^3 \theta}{3} \right) \cos \theta \, d\theta.$

Step 5: Simplify the Integral

Distribute $\cos \theta$ to the terms inside the integral: $\int_0^{\frac{\pi}{2}} \left( \frac{8 \cos \theta}{3} - \frac{8 \cos^4 \theta}{3} \right) \, d\theta.$

We now split this into two separate integrals: $\frac{8}{3} \int_0^{\frac{\pi}{2}} \cos \theta \, d\theta - \frac{8}{3} \int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta.$

Step 6: Evaluate the Integrals

1. The first integral is straightforward: $\int_0^{\frac{\pi}{2}} \cos \theta \, d\theta = \sin \theta \Big|_0^{\frac{\pi}{2}} = 1.$

So, the first term becomes: $\frac{8}{3} \times 1 = \frac{8}{3}.$

2. The second integral requires a standard reduction formula for $\cos^4 \theta$. Using the identity: $\cos^4 \theta = \frac{3}{8} + \frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta,$ we integrate each term: $\int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}.$

So, the second term becomes: $-\frac{8}{3} \times \frac{3\pi}{16} = -\frac{8 \pi}{16} = -\frac{\pi}{2}.$

Step 7: Combine the Results

Now, combine both terms: $\frac{8}{3} - \frac{\pi}{2}.$

Final Answer:

The value of the integral is: $\boxed{\frac{8}{3} - \frac{\pi}{2}}$

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