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How do you find the area of a region \(R=\\{(r, \theta): 0 \leq g(\theta) \leq r \leq h(\theta), \alpha \leq \theta \leq \beta\\} ?\)

Short Answer

Expert verified
Question: Given a polar region R with outer radius function h(θ) and inner radius function g(θ), and angle bounds α and β, find the area of the region R. Answer: Use the formula A = (1/2)∫(h(θ)^2 - g(θ)^2)dθ with limits α and β, where A is the area of the region R.

Step by step solution

01

Understand the area formula for a polar curve

The formula for finding the area of a polar region is given by: $$A = \frac{1}{2}\int_{\alpha}^{\beta}(h(\theta))^2 - (g(\theta))^2d\theta$$ This is derived from the fact that the area of a small sector of a polar curve can be approximated by a triangle with a base of length \(r\) and height \(rd\theta\). Integrating the difference of the squares of the outer and inner radius functions over the range of angles gives us the total area.
02

Identify the given radius functions and angle bounds

In the given exercise, the radius function for the inner boundary is \(g(\theta)\) and the radius function for the outer boundary is \(h(\theta)\). The angle bounds are given by \(\alpha \leq \theta \leq \beta\). We need to apply these functions and bounds to the area formula mentioned in Step 1.
03

Calculate the area using the given information

Now, we just need to plug in the given radius functions and angle bounds into the area formula: $$A = \frac{1}{2}\int_{\alpha}^{\beta}(h(\theta))^2 - (g(\theta))^2d\theta$$ By integrating with respect to θ, we can obtain the area of the region R. If the radius functions are given explicitly, substitute them in the formula and proceed with integration. If the radius functions are not given, then further information about the specific problem will be needed to determine the area. After evaluating the integral, the result will be the area of the given polar region R.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area of a Polar Region
To find the area of a region defined by polar coordinates, it's crucial to understand how polar curves work. Unlike Cartesian coordinates, where we use regular squares or rectangles, polar coordinates rely on sectors of a circle, shaped like slices of a pie. The area of a small sector is approximated by a triangle with a base of length \(r\) and height \(r\Delta \theta\).

The formula for finding the area of a polar region is given by:
  • \(A = \frac{1}{2}\int_{\alpha}^{\beta}((h(\theta))^2 - (g(\theta))^2)\,d\theta\)
Here's what each part represents:
  • \(\alpha\) and \(\beta\) are the angle bounds.
  • \(h(\theta)\) is the outer radius function.
  • \(g(\theta)\) is the inner radius function.
The formula calculates the area by subtracting the area enclosed by the inner curve from the area enclosed by the outer curve.
This way, you only get the area of the region between the two curves.
This subtraction accurately helps in capturing the space defined by these curves.
Calculating Integrals
Once the radius functions and angle bounds are identified, calculating the integral is the next step. The integral helps sum up infinitely small areas to find the total area of the region. In polar coordinates, the integral considers the slice-like nature of the areas, which makes it different from regular Cartesian integrals.

Here’s how you perform the integration:
  • Identify the outer and inner radius functions, \(h(\theta)\) and \(g(\theta)\).
  • Set your limits of integration between angles \(\alpha\) and \(\beta\).
  • The integrand is \((h(\theta))^2 - (g(\theta))^2\).
  • Integrate this function with respect to \(\theta\).
The integral will sum up all the small differential areas to give the total area.
Make sure to consider every part of the formula carefully while integrating.
Radius Functions
Radius functions in polar coordinates are crucial as they define the reach or extent of the curve from the center (origin). Each angle value \(\theta\) corresponds to a radius—either \(g(\theta)\) or \(h(\theta)\). These functions dictate the boundaries of the region whose area we want to find.

Key things to note about radius functions:
  • \(h(\theta)\) generally represents the outer boundary, extending further from the center.
  • \(g(\theta)\) is usually the inner boundary, closer to the center.
  • As \(\theta\) changes, these radii create the varying distances needed to define the polar region.
Understanding their behavior and range is essential to applying them properly in the area formula.
The main task is to substitute these functions into the integral, allowing you to accurately compute the polar region's area.

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Most popular questions from this chapter

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\begin{array}{l} \iint_{R} \frac{d A}{4+\sqrt{x^{2}+y^{2}}} ; R=\\{(r, \theta): 0 \leq r \leq 2 \\\ \pi / 2 \leq \theta \leq 3 \pi / 2\\} \end{array}$$

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 an ellipsoid with axes of length \(2 a\) \(2 b,\) and \(2 c\)

General volume formulas Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Frustum of a cone Find the volume of a truncated solid cone of height \(h\) whose ends have radii \(r\) and \(R\).

Open and closed boxes Consider the region \(R\) bounded by three pairs of parallel planes: \(a x+b y=0, a x+b y=1\) \(c x+d z=0, c x+d z=1, e y+f z=0,\) and \(e y+f z=1\) where \(a, b, c, d, e,\) and \(f\) are real numbers. For the purposes of evaluating triple integrals, when do these six planes bound a finite region? Carry out the following steps. a. Find three vectors \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) each of which is normal to one of the three pairs of planes. b. Show that the three normal vectors lie in a plane if their triple scalar product \(\mathbf{n}_{1} \cdot\left(\mathbf{n}_{2} \times \mathbf{n}_{3}\right)\) is zero. c. Show that the three normal vectors lie in a plane if ade \(+b c f=0\) d. Assuming \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) lie in a plane \(P,\) find a vector \(\mathbf{N}\) that is normal to \(P .\) Explain why a line in the direction of \(\mathbf{N}\) does not intersect any of the six planes and therefore the six planes do not form a bounded region. e. Consider the change of variables \(u=a x+b y, v=c x+d z\) \(w=e y+f z .\) Show that $$J(x, y, z)=\frac{\partial(u, v, w)}{\partial(x, y, z)}=-a d e-b c f$$ What is the value of the Jacobian if \(R\) is unbounded?

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 3} \int_{0}^{4 \sec \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

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