Question

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 the cap of a sphere of radius \(R\) with height \(h\)

Short answer

Answer: The volume formula for a spherical cap is given by V = 2π[((2/3)R^4 - (2/3)R^3(R-h) - Rh(R-h)^2 + (1/3)R(R-h)^3 - (1/4)(R-h)^4)].

Step-by-step solution

Identify the bounding surfaces

First, we need to identify the surfaces that define the cap. The cap is formed by cutting the sphere at a plane, which is parallel to the base and h units below the top of the sphere. The bounding surfaces are the sphere itself and the planar surface that cuts the sphere.

Sphere Equation

Let's first write down the equation for the sphere. The sphere equation is given by: (x^2 + y^2 + z^2 = R^2)

Planar Surface Equation

Now we need to find the equation of the planar surface cutting the sphere. To do this, we know that the hemisphere's highest point is at (0, 0, R), and our plane is h units below it. So a point on the plane will be (0, 0, R-h). Furthermore, the plane is parallel to the xy-plane, meaning the z-coordinate will remain constant at R-h. Therefore, the equation of the planar surface is given by: z = R - h

Set up the volume integral

The volume integral will be in the form of a double integral over the projection of the cap on the xy-plane. For simplicity, we will convert the Cartesian coordinates to polar coordinates. In this case x = r cos(θ) and y = r sin(θ), so the integral is over r and θ. The volume element in polar coordinates is drdθrdz. The volume integral is: V = ∬[∫(R - h - z) drdθ] rdz

Set up the integral limits

To set up the limits of integration, we need first to find the range of r in terms of z. Using the sphere equation (x^2 + y^2 + z^2 = R^2) and converting to polar coordinates, we get r^2 = R^2 - z^2. So, r ranges from 0 to √(R^2 - z^2). Then, we need to find the range of θ, which is simply from 0 to 2π, because the cap is a full circular section. Finally, z ranges from R-h to R, which are the limits of our sphere and our planar surface. Now we can write the integral with the limits: V = ∫[∫[∫(R - h - z) dr]dθ] dz where r ranges from 0 to √(R^2 - z^2), θ ranges from 0 to 2π, and z ranges from R-h to R.

Evaluate the integral

Now, we will evaluate the integral step by step: First, we will integrate with respect to r: V = ∫[∫[Rr - hr + zr] from r = 0 to r = √(R^2 - z^2)] dθdz Next, we will integrate with respect to θ: V = ∫[(2πRr - 2πhr + 2πzr) from θ = 0 to θ = 2π]dz V = ∫[2π(R^3 - Rh^2 + Rz^2 - z^3)]dz from z = R-h to z = R Finally, we will integrate with respect to z: V = [2π((2/3)R^3z - Rhz^2 + (1/3)Rz^3 - (1/4)z^4)] from z = R-h to z = R Substitute the limits, we get: V = 2π[((2/3)R^4 - (2/3)R^3(R-h) - Rh(R-h)^2 + (1/3)R(R-h)^3 - (1/4)(R-h)^4)] This is the volume formula for the cap of a sphere of radius R with height h.

Key concepts

Integration in Polar Coordinates

When tackling volume calculations, specifically for the cap of a sphere, it's efficient to use polar coordinates. This transformation simplifies the integration process. Here's why: In polar coordinates, every point in space is represented in terms of a radius and an angle rather than rectangular coordinates. You can convert the Cartesian coordinates \(x, y\) to the polar system using \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). The advantage here is that complex shapes like circular or spherical regions often become easier to describe and compute.
For volume integrals in polar coordinates, the area element becomes \(r \, dr \, d\theta \, dz\). This is due to the additional radial stretch factor \(r\) which needs to be included, and it reflects the change from rectangular area to a section of a circle.
This setup allows us to integrate over the circular projection in the \(xy\)-plane effortlessly.

Sphere Equation

Understanding the sphere equation is fundamental when dealing with spherical volumes. The standard equation for a sphere centered at the origin is given by \(x^2 + y^2 + z^2 = R^2\), where \(R\) is the radius.
This equation tells us that for any point \(x, y, z\) on the sphere, its radial distance from the origin is equal to \(R\). Visualize this as all the points equidistant from a central point, forming a perfectly balanced three-dimensional shape.
In the context of our problem, knowing the sphere's equation allows us to relate the variables when setting up the integral limits. As \(z\) varies within the specified limits, it helps determine the possible values (or bounds) for \(r\) and \(\theta\). It's an essential piece for framing both the area of integration and establishing the bounding surfaces.

Volume Integration

In volume integration, the objective is to find the accumulated volume under certain surfaces, which requires setting up correctly bounded integrals. The problem of finding the volume of a spherical cap emphasizes this method.
The integral is set as a triple integral here for the volume, starting by integrating over \(r\), then over \(\theta\), and finally over \(z\). The form of our integral is:

• Integrate the radial component \(r\) first, as \(0 \, \text{to} \, \sqrt{R^2 - z^2}\)
• Then the entire circle described by \(\theta\) ranging from \(0\) to \(2\pi\)
• Lastly, along the vertical axis \(z\), from \(R-h\) to \(R\)

The calculated integrals provide the volume below the cap intersected by the plane. The details of this integration handle the volume's layered build as it progresses through these regions. Each step hones in on reducing a specific dimension until the entire volume is determined.

Bounding Surfaces in Calculus

Bounding surfaces describe the limits and borders of the space where the integration occurs. In Calculus, these are the surfaces against which you're evaluating your integrals. Understanding them is crucial in volume problems.
For the spherical cap, two distinct bounding surfaces are considered: the sphere's surface defined by \(x^2 + y^2 + z^2 = R^2\), and the planar cut through the sphere given by \(z = R - h\). This intersection between the sphere and plane forms the cap.

• The sphere surface encompasses the outer boundary.
• The planar surface provides a truncating border for the cap.

These surfaces help establish the region of integration. They are key to understanding both what is being bounded and the integration limits that need to be used to solve such problems. Each creates a piece of the puzzle that limits the extent of the variables in your integrals.