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Use a graphing utility to graph the polar equation and find the area of the given region.Inner loop of \(r=4-6 \sin \theta\)

Short Answer

Expert verified
Firstly, understand the polar function and sketch the graph. Identify the inner loop, and use the area formula for polar coordinates to set up the integral. Subsequently compute the integral to obtain the area.

Step by step solution

01

Understand the polar function: \(r=4-6 \sin \theta\)

The graph of any function of the form \(r = a \pm b \sin \theta \) or \(r = a \pm b \cos \theta \) (where both 'a' and 'b' are positive) makes a rose curve. If 'a' is equal to 'b', then there is no inner loop. If 'a' is not equal to 'b', then there is a smaller loop called an 'inner loop'.
02

Graph the polar function

Graph the polar function using a graphing utility. Polar graphing involves the radial function 'r' which tells us how far from the origin to plot the point and the angle theta (\(\theta\)) which tells us the direction to that point. When graphing, notice that the 'inner loop' occurs when \(\theta\) is between \(-\pi/2\) and \(\pi/2\). Contrastingly, the 'outer loop' occurs when \(\theta\) is between \(\pi/2\) and \(-\pi/2\).
03

Find the area under the polar curve

Apply the formula for the area \(A\) under a polar curve \(r=f(\theta)\) from \(\theta=\alpha\) to \(\theta=\beta\): \(A=\frac{1}{2}\int_{\alpha}^{\beta}[f(\theta)]^2 d\theta\). Here, the integration limits for the inner loop are \(\theta=-\pi/2\) to \(\pi/2\). Hence, we will integrate \(r(\theta) = 4-6 sin\theta\) from \(-\pi/2\) to \(\pi/2\). The result will then be multiplied by \(\frac{1}{2}\).
04

Compute the definite integral

Let's compute the definite integral: \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(4-6 \sin \theta)^2 d\theta\). Carry out the algebra to square the integrand creating three terms, then compute the integral of each term separately. These will be simple integrals \(\int a d\theta = a\theta\) and \(\int \sin^2(\theta) d\theta\).
05

Evaluate the definite integral using Fundamental Theorem of Calculus

To evaluate the integral, subtract the antiderivative evaluated at the lower limit of integration \(-\pi/2\) from the antiderivative evaluated at the upper limit \(\pi/2\). The answer we get is the area of the region.

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Key Concepts

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

Graphing Polar Functions
Understanding how to graph polar functions is fundamental in mathematics, especially when considering the region they enclose. A polar function presents a relationship between the radius r, which is the distance from the origin to a point, and the angle θ, which determines the direction from the origin to that point. Unlike Cartesian coordinates that use a grid of x and y, polar graphing is done on a circular grid.

When graphing the polar function r=4-6 sin θ, you plot the points by calculating the radius for different angles. This will result in a 'rose curve,' and if the coefficients in the function are not equal, the graph will feature an inner loop. This loop is a smaller section of the curve, which can be graphed using technology like graphing calculators or software. For the area calculation, only the points where the curve has an inner loop are considered, which usually occurs within specific ranges of θ.
Integral Calculus
Integral calculus is a part of mathematics that deals with finding the area under a curve, volumes, and other related quantities. It is used to sum infinitely small factors to determine a whole.

In the context of polar equations, to find the area enclosed by a polar curve, you use a specific integral calculus formula. The beauty of integral calculus lies in its ability to handle these complex shapes and curves that cannot be easily broken down into simple geometries like triangles or rectangles. The integration process adapts to the curve's nature, considering every point - hence, you can calculate precise areas of these unique shapes.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus unites the two principal concepts of calculus: differentiation and integration. It states that if a function is continuous over an interval and has an antiderivative in that interval, the definite integral of the function over that interval is equal to the difference between the values of the antiderivative at the endpoints.

This theorem is crucial when we calculate the area under the polar curve. Once the integrand is set up, we evaluate it at the limits of integration. The evaluation gives us the exact area of the shape described by the function within these limits. Without this theorem, the connection between the antiderivative and the area under the curve wouldn't be as clear, and finding areas analytically would be a much harder task.
Polar Coordinates
Polar coordinates are an alternative to the Cartesian coordinate system most people are familiar with. Instead of using x and y to denote a point's location, we use a radius r and an angle θ. This makes it particularly useful for circles and shapes where symmetry about the origin plays a role.

Using polar coordinates can simplify equations for curves and shapes in the plane, especially when dealing with circular or spiral patterns. This change from Cartesian to polar often provides a more natural and simple description of a curve's properties, which is particularly evident when trying to understand the inner loops or petals of certain polar graphs.

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Most popular questions from this chapter

Folium of Descartes A curve called the folium of Descartes can be represented by the parametric equations \(x=\frac{3 t}{1+t^{3}} \quad\) and \(\quad y=\frac{3 t^{2}}{1+t^{3}} .\) (a) Convert the parametric equations to polar form. (b) Sketch the graph of the polar equation from part (a). (c) Use a graphing utility to approximate the area enclosed by the loop of the curve.

The path of a projectile is modeled by the parametric equations \(x=\left(90 \cos 30^{\circ}\right) t \quad\) and \(\quad y=\left(90 \sin 30^{\circ}\right) t-16 t^{2}\) where \(x\) and \(y\) are measured in feet. (a) Use a graphing utility to graph the path of the projectile. (b) Use a graphing utility to approximate the range of the projectile. (c) Use the integration capabilities of a graphing utility to approximate the arc length of the path. Compare this result with the range of the projectile.

The planets travel in elliptical orbits with the sun as a focus, as shown in the figure. (a) Show that the polar equation of the orbit is given by \(r=\frac{\left(1-e^{2}\right) a}{1-e \cos \theta}\) where \(e\) is the eccentricity. (b) Show that the minimum distance (perihelion) from the sun to the planet is \(r=a(1-e)\) and the maximum distance \((\) aphelion \()\) is \(r=a(1+e)\).

Use the formula for the arc length of a curve in parametric form to derive the formula for the arc length of a polar curve.

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. \(x=\sqrt{t}, \quad y=t-2\)

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