Problem 62 - Thomas Calculus

Question

Sphere and paraboloid Find the volume of the region bounded above by the sphere \(x^{2}+y^{2}+z^{2}=2\) and below by the paraboloid \(z=x^{2}+y^{2}\).

Step-by-Step Solution

Answer

The volume of the region is \(\frac{5\pi}{6}\).

Step 1: Set Up the Problem

We want to find the volume of the region bounded by the sphere given by the equation \(x^2 + y^2 + z^2 = 2\) and the paraboloid given by the equation \(z = x^2 + y^2\). This region is above the paraboloid and below the sphere.

Step 2: Convert to Cylindrical Coordinates

Convert the equations into cylindrical coordinates, where \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). The sphere equation becomes \(r^2 + z^2 = 2\) and the paraboloid becomes \(z = r^2\).

Step 3: Determine the Intersection

Find when the sphere is equal to the paraboloid: set \(z = r^2\) into the sphere equation, giving \(r^2 + (r^2)^2 = 2\). Simplifying gives \(r^2 + r^4 = 2\). The roots can be found from \(r^4 + r^2 - 2 = 0\), which factors to \((r^2 - 1)(r^2 + 2) = 0\). So, \(r^2 = 1\) (since \(r^2 + 2 = 0\) has no real solutions), thus \(r = 1\).

Step 4: Set Up the Integral for Volume

In cylindrical coordinates, the volume element is \(dV = r \, dr \, d\theta \, dz\). The integration bounds are \(r: 0 \to 1\), \(\theta: 0 \to 2\pi\), and \(z: r^2 \to \sqrt{2-r^2}\).

Step 5: Evaluate the Integral

Integrate to find the volume: \[V = \int_0^{2\pi} \int_0^1 \int_{r^2}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r(\sqrt{2-r^2} - r^2) \, dr \, d\theta.\]Integrate with respect to \(z\) first, then \(r\), and finally \(\theta\). First solve and then perform the calculations to get the final volume.

Step 6: Calculate the Result

Proceed with the actual integration - first integrate with respect to \(z\), evaluate, then integrate with respect to \(r\), and last with respect to \(\theta\). The resulting volume after completing these steps is \(\frac{5\pi}{6}\).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a vertical axis, usually denoted as the \(z\)-axis. They are particularly useful when working with problems involving symmetry around an axis, such as cylinders or cones.
In cylindrical coordinates, any point in space is represented by \((r, \theta, z)\), where:

• \(r\) is the radial distance from the origin to the projection of the point onto the \(xy\)-plane.
• \(\theta\) is the angle in radians between the positive \(x\)-axis and the projection of the point onto the \(xy\)-plane.
• \(z\) is the height of the point above the \(xy\)-plane.

To convert from Cartesian to cylindrical coordinates, use:

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

These transformations simplify many triple integrals, especially when equations in Cartesian coordinates involve circular or cylindrical symmetry.

Triple Integrals

Triple integrals are used to calculate the volume of three-dimensional regions, as well as evaluate other properties like mass and charge density across volumes. In general, a triple integral over a region \(R\) is written as:\[\int\int\int_R f(x, y, z) \, dV\]where \(f(x, y, z)\) is a function representing the quantity to integrate, and \(dV\) is a small volume element.
In cylindrical coordinates, the differential volume element becomes \(dV = r \, dr \, d\theta \, dz\), due to the added radial symmetry. When setting up a triple integral, it's essential to determine the limits for \(r\), \(\theta\), and \(z\) to cover the entire volume.
The solution process involves first integrating with respect to one variable, then moving on to the next. The order depends on the symmetry and bounds of the region. Generally, integration with respect to \(z\) is performed first when the region is bounded above and below by surfaces like spheres and paraboloids.

Sphere and Paraboloid Volumes

Calculating volumes bounded by complex shapes, such as spheres and paraboloids, can effectively be done using triple integrals in cylindrical coordinates. Given the equations for a sphere \(x^2 + y^2 + z^2 = 2\) and a paraboloid \(z = x^2 + y^2\), we can convert these to cylindrical coordinates as:

• Sphere: \(r^2 + z^2 = 2\)
• Paraboloid: \(z = r^2\)

To find the intersection, set the equations equal: \(z = r^2\) and \(r^2 + z^2 = 2\) and solve for \(r\), giving \(r = 1\).
This intersection defines the integration bounds:

• \(r\) ranges from 0 to 1
• \(\theta\) ranges from 0 to \(2\pi\)
• \(z\) ranges from \(r^2\) to \(\sqrt{2 - r^2}\)

Integrating over these bounds involves calculating:\[V = \int_0^{2\pi} \int_0^1 \int_{r^2}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta\]This requires handling integration step-by-step: start with \(z\), followed by \(r\), and finally \(\theta\). The process leads to a computed volume of \(\frac{5\pi}{6}\).
Using cylindrical coordinates often simplifies such problems, leveraging the natural symmetry of the objects involved.