Problem 47 - Thomas Calculus

Question

Find the volumes of the solids. \(z=\sqrt{1-x^{2}-y^{2}}\) \(r=\sin \theta\)

Step-by-Step Solution

Answer

The volume of the solid is the integral's evaluated value, solved through cylindrical coordinates and trigonometric integration.

Step 1: Understand the Given Equations

We have two surfaces defined by equations. The first equation is the upper half of a sphere given by \( z = \sqrt{1 - x^2 - y^2} \), which is a sphere with radius 1 centered at the origin. The second equation \( r = \sin \theta \) represents a circle in the polar coordinate system.

Step 2: Convert to Cylindrical Coordinates

To find the volume using the given surfaces, it's helpful to use cylindrical coordinates where \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = z \). In cylindrical coordinates, the surface \( z = \sqrt{1 - x^2 - y^2} \) becomes \( z = \sqrt{1 - r^2} \).

Step 3: Determine the Limits of Integration

The equation \( r = \sin \theta \) describes the boundary in the xy-plane. For full spherical symmetry (\(0 \leq \theta \leq 2\pi\), \(0 \leq r \leq 1\)), the intersection creates a circular region. The z-component ranges vertically from 0 to \( \sqrt{1 - r^2} \).

Step 4: Set Up the Integral for Volume

The volume \( V \) in cylindrical coordinates can be expressed as: \[ V = \int_0^{2\pi} \int_0^{\sin \theta} \int_0^{\sqrt{1-r^2}} r\, dz\, dr\, d\theta \]Here, \(r\) is included in the integral to convert the area element in cylindrical coordinates.

Step 5: Evaluate the Integral

Start by integrating with respect to \( z \):\[ \int_0^{\sqrt{1 - r^2}} r\, dz = r[z]_0^{\sqrt{1-r^2}} = r \sqrt{1-r^2} \]Next, integrate with respect to \( r \):\[ \int_0^{\sin \theta} r \sqrt{1 - r^2}\, dr \]Using substitution, let \( u = 1 - r^2 \), \( du = -2r\, dr \) and change the limits of \( r \). Finally, integrate with respect to \( \theta \) from 0 to \( 2\pi \).

Step 6: Compute the Final Result

After computing the integrals, we get the volume as a number. Solving the integrals through substitution and simplification, confirm the region of integration and execute the definite integrals for precise number. Substitute back any trigonometric identities and evaluate the limits for\( \theta \).

Key Concepts

cylindrical coordinates

Cylindrical coordinates extend the concept of polar coordinates into three dimensions. They are particularly useful when dealing with solids of revolution and cylindrical symmetry. In cylindrical coordinates, we describe a point in space with three components: radial distance \( r \), angle \( \theta \), and height \( z \). This system is useful when symmetry about an axis (often the z-axis) is present, simplifying computations for volume and surface integrals.

When solving for volumes, such as in this problem, the conversion from Cartesian to cylindrical coordinates involves the transformations:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( z = z \)

These transformations simplify integration over circular regions and volumes, as the radial and angular components relate directly to the geometry of the problem. In this exercise, the volume integral is constructed in cylindrical coordinates to take advantage of the axial symmetry of the sphere described by \( z = \sqrt{1 - r^2} \). This simplification reduces the need to split the geometry into separate integrals associated with Cartesian limits.

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. They are defined by:

• \( r \): radial distance from the origin
• \( \theta \): angle from the positive x-axis

Polar coordinates are ideal for dealing with problems involving circular or rotational symmetry.

In this context, the problem specifies a curve in polar coordinates: \( r = \sin \theta \). This curve describes a circle centered at the origin with a radius reaching outwards in the polar plane. Transformations from polar to Cartesian can be made with the formulas:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

This relationship is essential when relating the conditions of the spherical cap to circular boundaries on the xy-plane. Polar coordinates, therefore, provide a straightforward method to set up integration limits for circular regions in problems involving integrals over circular or annular areas.

integration limits

Integration limits define the region over which integration occurs. They vary according to the coordinate system used. For cylindrical coordinates, we establish limits for \( r \), \( \theta \), and \( z \) separately. In this problem, the integration occurs in three dimensions with ranges specified to capture the volume under the surface \( z = \sqrt{1 - r^2} \).

The integration structure in this problem is:

• \( \theta \) ranges from 0 to \( 2\pi \) to complete a full rotation around the z-axis.
• \( r \) is bounded by the polar curve \( r = \sin \theta \), creating a circular boundary in the xy-plane.
• \( z \) is vertically constrained from 0 up to the height specified by \( z = \sqrt{1-r^2} \), corresponding to the upper half-sphere.

This setup ensures integration over the desired 3D region, capturing the entire volume under the spherical surface intersected by the plane described in polar form.