Problem 48 - Calculus Early Transcendentals

Question

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the cardioid \(r=3-3 \cos \theta\)

Step-by-Step Solution

Answer

The coordinates of the centroid of the region bounded by the cardioid \(r = 3 - 3\cos\theta\) are \((\bar{x}, \bar{y}) = (0, \frac{2}{3})\).

Step 1: Find the area of the region

To find the area of the region bounded by the cardioid \(r=3-3 \cos \theta\), we integrate with respect to \(\theta\) using the following formula: $$A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$$ In this case, our entire region covers \(0 \leq \theta \leq 2\pi\). Therefore, the area of the region is: $$A = \frac{1}{2} \int_{0}^{2\pi} (3-3\cos\theta)^2 d\theta$$

Step 2: Evaluate the area integral

We now need to evaluate the integral to find the area of the region. Let's expand the square term: $$A = \frac{1}{2} \int_{0}^{2\pi} (9-18\cos\theta +9\cos^2\theta) d\theta$$ To simplify the integral, use the double angle identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\): $$A = \frac{1}{2} \int_{0}^{2\pi} (9-18\cos\theta +\frac{9 + 9\cos 2\theta}{2}) d\theta$$ Now, we can integrate term-by-term: $$A = \frac{1}{2} \bigg[ 9\theta -18\sin\theta +\frac{9}{2}\theta + \frac{9}{4}\sin(2\theta) \bigg]_0^{2\pi}$$ We can now evaluate the expression to get the area: $$A = \frac{1}{2} \bigg[ 18\pi - 0 + 9\pi - 0 \bigg] = \frac{27}{2}\pi$$

Step 3: Find the centroid coordinates

To find the centroid coordinates \(\bar{x}\) and \(\bar{y}\), we need to calculate the average x and y coordinates and divide by the area A. We can use the following formulas for this: $$\bar{x} = \frac{1}{A} \int_{\alpha}^{\beta}{r\cos\theta r d\theta}= \frac{1}{27\pi} \int_{0}^{2\pi}{(3-3\cos\theta)\cos\theta(3-3\cos\theta) d\theta}$$ $$\bar{y} = \frac{1}{A} \int_{\alpha}^{\beta}{r\sin\theta r d\theta}= \frac{1}{27\pi} \int_{0}^{2\pi}{(3-3\cos\theta)\sin\theta(3-3\cos\theta) d\theta}$$

Step 4: Evaluate the integrals for \(\bar{x}\) and \(\bar{y}\)

Now, let's integrate each expression for \(\bar{x}\) and \(\bar{y}\): $$\bar{x} = \frac{1}{27\pi} \int_{0}^{2\pi}{(9-18\cos\theta + 9\cos^2\theta)\cos\theta d\theta}$$ Since \(\cos\theta\) is an odd function, any even power multiplied by the cosine will result in an odd function, and since the integral limits are symmetric, the integral becomes zero. $$\bar{x} = 0$$ Now, let's evaluate the integral for \(\bar{y}\): $$\bar{y} = \frac{1}{27\pi} \int_{0}^{2\pi}{(9-18\cos\theta + 9\cos^2\theta)\sin\theta d\theta}$$ Using integration by parts, we find that: $$\bar{y} = \frac{1}{27\pi} \bigg[ 9\cos\theta +18\sin\theta -9\frac{\sin\theta\cos\theta}{2} \bigg]_0^{2\pi}= \frac{18}{27} = \frac{2}{3}$$

Step 5: Write the final result

The centroid of the region bounded by the cardioid \(r=3-3 \cos \theta\) is at point \((\bar{x}, \bar{y})=(0, \frac{2}{3})\).