Question

Use cylindrical coordinates to find the indicated quantity. Volume of the solid bounded above by the sphere \(r^{2}+z^{2}=5\) and below by the paraboloid \(r^{2}=4 z\)

Short answer

The volume of the solid is given by the integral \(\int_{0}^{2\pi} \int_{0}^{2} r(\sqrt{5 - r^2} - \frac{r^2}{4}) \, dr \, d\theta\).

Step-by-step solution

Understanding the Problem

We are given a volume bounded by a sphere and a paraboloid in cylindrical coordinates. The top boundary is given by the sphere equation \(r^2 + z^2 = 5\), and the bottom boundary is given by the paraboloid \(r^2 = 4z\). We need to find the region of volume by integrating using cylindrical coordinates \((r, \theta, z)\).

Setting up the Integration Limits

First, we express the intersection of the sphere and paraboloid by equating their equations: \(r^2 + z^2 = 5\) and \(r^2 = 4z\). Substituting \(r^2 = 4z\) into the sphere's equation gives \((4z) + z^2 = 5\), which simplifies to \(z^2 + 4z - 5 = 0\). Solving this quadratic equation, we find the roots \(z = 1\) and \(z = -5\). Since \(z\) must be non-negative, we use \(z=1\). The integration limits for \(z\) are from the paraboloid \(z = \frac{r^2}{4}\) to the sphere \(z = \sqrt{5 - r^2}\).

Determine Radial and Angular Limits

For \(r\), as \(z\) goes from \(z = 0\) to \(z = 1\), \(r\) should vary such that it is determined by the sphere at the intersection, so it's from \(0\) to \(2\), solved from \(z=1\) in \(r^2 = 4z\) which gives \(r=2\). The angle \(\theta\) varies full circle from \(0\) to \(2\pi\).

Set Up the Integral

The volume \(V\) can be set up as a triple integral in cylindrical coordinates. The integral is given by: \[ V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2/4}^{\sqrt{5 - r^2}} r \, dz \, dr \, d\theta \].

Evaluate the Inner Integral

Integrate with respect to \(z\):\[ \int_{r^2/4}^{\sqrt{5 - r^2}} r \, dz = r[z]_{r^2/4}^{\sqrt{5 - r^2}} = r(\sqrt{5 - r^2} - \frac{r^2}{4}) \].

Integrate Over \(r\) and \(\theta\)

Now, perform the integration over \(r\) and \(\theta\):\[ V = \int_{0}^{2\pi} \int_{0}^{2} r(\sqrt{5 - r^2} - \frac{r^2}{4}) \, dr \, d\theta \].Evaluating the definite integrals, first over \(r\), and then over \(\theta\), would yield the final volume.

Key concepts

Volume Calculation

In calculus, finding the volume of complex shapes is often achieved by using integration. For shapes not bound by straight edges, like those bounded by spheres and paraboloids, cylindrical coordinates become particularly useful. The process begins by visualizing the shape in the correct coordinate system so that we can set up the right kind of integrals to solve for volume. In this exercise, the solid lies between a sphere and a paraboloid. By understanding their geometry, we can establish the limits and integrals necessary for finding the volume.

• We translate the geometric boundaries into mathematical equations.
• Using cylindrical coordinates helps because it aligns with rotational symmetry commonly found in spheres and paraboloids.
• The volume calculation translates into a triple integral.

The actual volume is calculated by integrating the space contained between these surfaces, dissected in radial, angular, and vertical components. This setup allows us to calculate volumes that standard geometry wouldn’t easily solve for.

Integration Limits

Setting integration limits is crucial to evaluate the integral accurately. These limits define the space we are integrating over.
The correct limits of integration ensure that the entire intended volume is captured, and none of it extends into non-relevant space.

• The radial boundary, described by equations like \(r^2 = 4z\), defines the limit for the radius \(r\) in terms of height \(z\).
• The upper boundary \(z = \sqrt{5-r^2}\) comes from the sphere equation and sets the limit along the vertical axis \(z\).
• Angular limits in cylindrical coordinates generally span from \(0\) to \(2\pi\), reflecting the full rotation around the z-axis.

For this problem, recognizing where these boundaries meet each other shows where volume calculation begins and ends in the three-dimensional space.

Triple Integral

A triple integral expands the concept of ordinary integration by integrating over three dimensions instead of one. In cylindrical coordinates, this approach tackles the radial, angular, and vertical components in a sequential manner.
To solve the problem, the initial setup of the integral is crucial.

• The \(r \, dz\, dr\, d\theta\) format signifies integration over volume.
• First, integrate with respect to \(z\) between its specific limits which varies with \(r\).
• Next, integrate radially over \(r\), capturing the volume's spread from the center outward.
• Finally, integration over \(\theta\) puts the volume calculation into a full circle, encapsulating all radial slices.

This ordered integration method ensures that every slice of the solid, bound by the paraboloid and sphere, is accurately accounted for, thereby ensuring the entire volume is calculated.

Coordinate Geometry

Understanding coordinate geometry is pivotal to transitioning between different coordinate systems like Cartesian and cylindrical, which can simplify the integration process. Cylindrical coordinates are specifically useful for problems involving rotational symmetry around a central axis, as seen in this exercise.

• In cylindrical coordinates \(r, \theta, z\), \(r\) represents the radial distance from the origin. It replaces the Cartesian \((x, y)\) plane with circular symmetry.
• The angle \(\theta\) provides orientation in the \((x, y)\) plane around the z-axis, representing the 360-degree rotational capacity.
• The vertical component \(z\) remains unchanged, similar to its Cartesian counterpart.

This coordinate system simplifies the math involved in problems with circular boundaries, reducing complex 3D surfaces into more manageable equations by exploiting inherent symmetries in the geometry. Recognizing the power of coordinate geometry in this way is fundamental for simplifying and solving integral calculus problems effectively.