Find the area of the part of the sphere of radius \(a\) and center at the origin which is above the square in the \((x, y)\) plane bounded by \(x=\pm a / \sqrt{2}\) and \(y=\pm a / \sqrt{2} .\) Hint for evaluating the integral: Change to polar coordinates and evaluate the \(r\) integral first.

Short Answer

Expert verified
Following the steps, we use the hint to change to polar coordinates and integrate over the appropriate bounds.

Step by step solution

01

- Understand the problem

Analyze the problem of finding the area of a part of the sphere with radius \(a\) and center at the origin. The part of the sphere lies above a square in the \(xy\)-plane bounded by \(x = \pm \frac{a}{\sqrt{2}}\) and \(y = \pm \frac{a}{\sqrt{2}}\).
02

- Setting up the integral

To solve the problem, we will use surface area integration in spherical coordinates. The surface area element in spherical coordinates is given by \(dS = a \sin(\theta)d\theta d\phi\).
03

- Converting to polar coordinates

Use the hint to change the cartesian coordinates in the \(x, y\)-plane to polar coordinates. The bounds for the given square are:

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Key Concepts

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

spherical coordinates
Spherical coordinates are a system of curvilinear coordinates that are natural for describing positions on a sphere. Instead of using \(x, y, z\) Cartesian coordinates, spherical coordinates use three values: radius \(r\), polar angle \(\theta\) (also known as the colatitude), and azimuthal angle \(\theta\) (also referred to as the longitude).
The radius is the distance from the origin to the point, the polar angle measures the angle from the positive \(z\)-axis, and the azimuthal angle measures the angle in the \(xy\)-plane from the positive \(x\)-axis.

Using these coordinates, the surface element for a sphere of radius \(a\) is given by:
\[ dS = a^2 \sin(\theta) \ d\theta \ d\phi \]
This equation accounts for the curvature of the sphere when integrating over its surface.

To visualize spherical coordinates, imagine drawing a line from the center of the sphere to any point on its surface. The length of this line is the radius. The angle between this line and the positive \(z\)-axis is the polar angle \(\theta\). Finally, the angle between the projection of this line onto the \(xy\)-plane and the positive \(x\)-axis is the azimuthal angle \(\theta\).

polar coordinates
Polar coordinates are a two-dimensional coordinate system where each point in the plane is determined by a distance from a reference point and an angle from a reference direction. These coordinates are particularly useful in problems involving circular symmetry.
The reference point is called the pole (analogous to the origin in Cartesian coordinates), and the reference direction is typically the positive \(x\)-axis.

In polar coordinates, any point is identified by \(r \) and \(\theta \). The value \(|r| \) is the distance from the pole, and \(\theta \) is the angle from the positive \(x \)-axis.
To convert Cartesian coordinates \( (x, y ) \) to polar coordinates, we use:
  • \[ r = \sqrt{x^2 + y^2} \ ]
  • \[\theta = \text{arctan} \left(\frac{y}{x}\right) \]

Using polar coordinates simplifies integration over circular regions. When converting the given exercise's square region to polar coordinates, the bounds for \(\theta \) range from \[ \theta = -\frac{\pi}{4} \ to \ \frac{\theta}{4} \], and for \( r \) range from \ 0 \ to \ \frac{a}{\root{2}}\]

surface area element
The surface area element is an infinitesimal piece of surface area used in surface integrals. In traditional Cartesian coordinates, the surface area element is expressed differently based on the surface. For example, on a plane, it might be \( dA = dx \ dy \).

In spherical coordinates, which are suitable for surfaces like spheres, the surface area element has to account for the curvature of the surface. For a sphere of radius \( a \), positioned at the origin, this becomes:
\

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Most popular questions from this chapter

(a) Write a triple integral in cylindrical coordinates for the volume of the part of a ball between two parallel planes which intersect the ball. (b) Evaluate the integral in (a). Warning hint: Do the \(r\) and \(\theta\) integrals first. (c) Find the centroid of this volume.

A triangular lamina is bounded by the coordinate axes and the line \(x+y=6\). Find its mass if its density at each point \(P\) is proportional to the square of the distance from the origin to \(P\).

Verify each of the following answers for an indefinite integral by one or more of the methods suggested above. \(\int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\sinh ^{-1} \frac{x}{a} \quad\) or \(\quad \ln (x+\sqrt{x^{2}+a^{2}}) . \quad\) Hint: To find the sinh \(^{-1}\) form, make the substitution \(x=a \sinh u\). Or see Chapter 2, Sections 15 and 17.

The following notation is used in the problems: \(M=\) mass, \(\bar{x}, \bar{y}, \bar{z}=\) coordinates of center of mass (or centroid if the density is constant), \(I=\) moment of inertia (about axis stated), \(I_{x}, I_{y}, I_{z}=\) moments of inertia about \(x, y, z\) axes, \(I_{m}=\) moment of inertia (about axis stated) through the center of mass. Note: It is customary to give answers for \(I, I_{m}, I_{x},\) etc., as multiples of \(M\) (for example, \(I=\frac{1}{3} M l^{2}\) ). A rectangular lamina has vertices (0,0),(0,2),(3,0),(3,2) and density \(x y .\) Find (a) \(M\), (b) \(\bar{x}, \bar{y}\), (c) \(I_{x}, I_{y}\), (d) \( I_{m}\) about an axis parallel to the \(z\) axis. Hint: Use the parallel axis theorem and the perpendicular axis theorem.

As needed, use a computer to plot graphs and to check values of integrals. Find the centroid of the first quadrant part of the arc \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3} .\) Hint: Let \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\).

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