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Chapter 13: Problem 48

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

Short Answer

Expert verified
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 by step solution

01

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: $$A = \frac{1}{2} \int_{0}^{2\pi} (3-3\cos\theta)^2 d\theta$$
02

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: $$A = \frac{1}{2} \bigg[ 18\pi - 0 + 9\pi - 0 \bigg] = \frac{27}{2}\pi$$
03

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: $$\bar{x} = \frac{1}{A} \int_{\alpha}^{\beta}{r\cos\theta r d\theta}= \frac{1}{27\pi} \int_{0}^{2\pi}{(3-3\cos\theta)\cos\theta(3-3\cos\theta) d\theta}$$ $$\bar{y} = \frac{1}{A} \int_{\alpha}^{\beta}{r\sin\theta r d\theta}= \frac{1}{27\pi} \int_{0}^{2\pi}{(3-3\cos\theta)\sin\theta(3-3\cos\theta) d\theta}$$
04

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: $$\bar{y} = \frac{1}{27\pi} \bigg[ 9\cos\theta +18\sin\theta -9\frac{\sin\theta\cos\theta}{2} \bigg]_0^{2\pi}= \frac{18}{27} = \frac{2}{3}$$
05

Write the final result

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

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Most popular questions from this chapter

Integrals in strips Consider the integral $$I=\iint_{R} \frac{d A}{\left(1+x^{2}+y^{2}\right)^{2}}$$ where \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq a\\}\) a. Evaluate \(I\) for \(a=1 .\) (Hint: Use polar coordinates.) b. Evaluate \(I\) for arbitrary \(a > 0\) c. Let \(a \rightarrow \infty\) in part (b) to find \(I\) over the infinite strip \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y < \infty\\}\)

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid is enclosed by a hemisphere of radius \(a\). How far from the base is the center of mass?

Use integration to show that the circles \(r=2 a \cos \theta\) and \(r=2 a \sin \theta\) have the same area, which is \(\pi a^{2}\)

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by one leaf of the rose \(r=\sin 2 \theta,\) for \(0 \leq \theta \leq \pi / 2\) \((\bar{x}, \bar{y})=\left(\frac{128}{105 \pi}, \frac{128}{105 \pi}\right)$$(\bar{x}, \bar{y})=\left(\frac{17}{18}, 0\right)\)

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a tetrahedron whose vertices are located at \((0,0,0),(a, 0,0),(0, b, 0),\) and \((0,0, c)\)
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