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Give the integral formulas for the area of the surface of revolution formed when the graph of \(r=f(\theta)\) is revolved about (a) the \(x\) -axis and (b) the \(y\) -axis.

Short Answer

Expert verified
Integral formulas for the area of the surface of revolution when function \(r=f(\theta)\) is revolved about the x-axis is \(S_x = 2\pi\int_a^bf(\theta)\sin(\theta)\sqrt{1+(\frac{dy}{dx})^2}dx\) and when revolved about the y-axis is \(S_y = 2\pi\int_a^bf(\theta)\cos(\theta)\sqrt{1+(\frac{dx}{dy})^2}dy\).

Step by step solution

01

Understanding Polar Coordinates

Consider \(r = f(\theta)\) to be a function in polar coordinates, where \(r\) is the radial distance from the origin, and \(\theta\) is the angle measured in the anticlockwise direction from the positive x-axis. The Cartesian coordinates can be related to polar coordinates as, \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). So, \(r = f(\theta) = f(x / \sqrt{x^2 + y^2})\).
02

Formulating Revolving Around the x-axis

When revolving around the x-axis, we get a surface of revolution formed. Its area can be calculated using the integral formula. The integral formula for the area \(S\) of a surface of revolution around the x-axis is: \(S_x = 2\pi\int_a^by\sqrt{1+(\frac{dy}{dx})^2}dx\). Plug in the polar function \(y = r\sin(\theta) = f(\theta)\sin(\theta)\) to get the final formula for \(S_x\).
03

Formulating Revolving Around the y-axis

Similarly, when revolving around the y-axis, we attain another surface of revolution. Its area can again be calculated using the integral formula. The integral formula for the area \(S\) of a surface of revolution around the y-axis is: \(S_y = 2\pi\int_a^bx\sqrt{1+(\frac{dx}{dy})^2}dy\). Plug in the polar function \(x = r\cos(\theta) = f(\theta)\cos(\theta)\) to get the final formula for \(S_y\).

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Key Concepts

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

Understanding Polar Coordinates
Polar coordinates are a way of representing points in a plane using a distance from a reference point and an angle from a reference direction. Imagine standing at the center of a circle (the origin), and facing a certain direction (the positive x-axis). The length of any straight line you draw from your feet to the edge of the circle is the radial coordinate, denoted by \( r \), and the angle you turn from your original facing direction is the angular coordinate, denoted by \( \theta \).

In simple terms, each point in the plane is described by how far away it is from the origin and by the angle it subtends from the positive x-axis. The equations that translate polar coordinates to regular Cartesian coordinates are \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). For a given function \( r = f(\theta) \), this establishes a relationship where the value of \( r \) varies as \( \theta \) changes, creating a path or curve in the plane.
Integral Formula for Surface Area
The integral formula is central to calculating the surface area of a shape formed by rotating a curve around an axis — termed a surface of revolution. When you revolve a curve, it spins around a specified line creating a three-dimensional surface.

To find the area of this surface, we use calculus, specifically integration. The integral formula gives us the sum of infinite infinitesimally small surface elements to compute the entire area. For surfaces of revolution, this requires expressions that incorporate derivatives, as these represent slopes and are crucial for understanding curves as they rotate.
Revolving Around the x-axis
When we revolve a curve around the x-axis, we visualize the curve spinning around this horizontal line, effectively creating a three-dimensional surface or shape. For example, think of spinning a circle around a horizontal rod to form a doughnut shape.

The formula used to calculate this surface area when a polar graph \( r = f(\theta) \) is revolved around the x-axis is:
  • \( S_x = 2\pi\int_a^b y\sqrt{1+(\frac{dy}{dx})^2}dx \)
For polar functions, \( y = f(\theta)\sin(\theta) \) replaces \( y \). Integration over this expression gives the exact surface area of the 3D shape formed by this revolution.
Revolving Around the y-axis
In contrast, revolving a curve around the y-axis involves considering the curve's rotation about a vertical line. This is akin to spinning a semi-circle vertically to create a tall vase-like shape.

The integral formula needed to find the surface area for a polar curve \( r = f(\theta) \) that revolves around the y-axis is:
  • \( S_y = 2 \pi \int_a^b x \sqrt{1 + (\frac{dx}{dy})^2} dy \)
With polar coordinates, \( x = f(\theta)\cos(\theta) \) acts as \( x \) in this formula. Using integration with these substitutions gives the complete surface area generated by the rotation.

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Most popular questions from this chapter

Consider the parametric equations \(x=4 \cot \theta\) and \(y=4 \sin ^{2} \theta, \quad-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). (a) Use a graphing utility to graph the curve represented by the parametric equations. (b) Use a graphing utility to find the points of horizontal tangency to the curve. (c) Use the integration capabilities of a graphing utility to approximate the arc length over the interval \(\pi / 4 \leq \theta \leq \pi / 2\)

Sketch a graph of the polar equation. $$ r \equiv 2 $$

Consider a projectile launched at a height \(h\) feet above the ground and at an angle \(\theta\) with the horizontal. If the initial velocity is \(v_{0}\) feet per second, the path of the projectile is modeled by the parametric equations \(x=\left(v_{0} \cos \theta\right) t\) and \(y=h+\left(v_{0} \sin \theta\right) t-16 t^{2}\). The center field fence in a ballpark is 10 feet high and 400 feet from home plate. The ball is hit 3 feet above the ground. It leaves the bat at an angle of \(\theta\) degrees with the horizontal at a speed of 100 miles per hour (see figure). (a) Write a set of parametric equations for the path of the ball. (b) Use a graphing utility to graph the path of the ball when \(\theta=15^{\circ} .\) Is the hit a home run? (c) Use a graphing utility to graph the path of the ball when \(\theta=23^{\circ} .\) Is the hit a home run? (d) Find the minimum angle at which the ball must leave the bat in order for the hit to be a home run.

Find the arc length of the curve on the interval \([0,2 \pi]\). Hypocycloid perimeter: \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\)

(a) Use a graphing utility to graph each set of parametric equations. \(x=t-\sin t \quad x=2 t-\sin (2 t)\) \(y=1-\cos t \quad y=1-\cos (2 t)\) \(0 \leq t \leq 2 \pi \quad 0 \leq t \leq \pi\) (b) Compare the graphs of the two sets of parametric equations in part (a). If the curve represents the motion of a particle and \(t\) is time, what can you infer about the average speeds of the particle on the paths represented by the two sets of parametric equations? (c) Without graphing the curve, determine the time required for a particle to traverse the same path as in parts (a) and (b) if the path is modeled by \(x=\frac{1}{2} t-\sin \left(\frac{1}{2} t\right) \quad\) and \(\quad y=1-\cos \left(\frac{1}{2} t\right)\)

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