Problem 85 - 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 given by a numerical result from the integral of the bounded region.

Step 1: Identify the region

First, determine where the sphere and the paraboloid intersect. Set the equations equal to each other:\(x^2 + y^2 + z^2 = 2\) and \(z = x^2 + y^2\). Substitute \(z\) from the paraboloid equation into the sphere equation:\[x^2 + y^2 + (x^2 + y^2)^2 = 2\].

Step 2: Simplify and solve for z

Simplify the combined equation from Step 1:\[x^2 + y^2 + (x^2 + y^2)^2 = 2\] becomes \[(x^2 + y^2) + (x^2 + y^2)^2 = 2\].Let \(u = x^2 + y^2\). Then \(u + u^2 = 2\), which can be solved as a quadratic equation, \(u^2 + u - 2 = 0\).

Step 3: Solve the quadratic equation

Solve \(u^2 + u - 2 = 0\) using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), \(c = -2\):\[u = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}\].The solutions are \(u = 1\) and \(u = -2\). Since \(u = x^2 + y^2\), it must be non-negative, so \(u = 1\).

Step 4: Set integration limits

The volume integral is over the region bounded by \(0 \leq x^2 + y^2 \leq 1\). Convert to cylindrical coordinates, where \(x^2 + y^2 = r^2\), so \(0 \leq r^2 \leq 1\) thus \(0 \leq r \leq 1\) and \(0 \leq \theta < 2\pi\). The height is from the paraboloid to the sphere, so \(z = r^2\) to \(\sqrt{2-r^2}\).

Step 5: Set up the integral

The volume \(V\) is given by the triple integral in cylindrical coordinates:\[V = \int_0^{2\pi} \int_0^1 \int_{r^2}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta\].

Step 6: Evaluate the innermost integral

Compute the inner integral with respect to \(z\):\[\int_{r^2}^{\sqrt{2-r^2}} r \, dz = r[z]_{r^2}^{\sqrt{2-r^2}} = r(\sqrt{2-r^2} - r^2)\].

Step 7: Evaluate the radial integral

Evaluate the radial integral:\[\int_0^1 r(\sqrt{2-r^2} - r^2) \, dr \].This splits into:\[\int_0^1 r\sqrt{2-r^2} \, dr - \int_0^1 r^3 \, dr\].Use the substitution method for \(\int r\sqrt{2-r^2} \, dr\), \(u = 2-r^2\), leading to:\[\int_0^1 r\sqrt{2-r^2} \, dr = \int_2^1 \sqrt{u} \, (-\frac{1}{2})du\].

Step 8: Solve integrals and finalize

Solve the integrals from Step 7:\[\int_1^2 \sqrt{u} \, du = \left[\frac{2}{3}u^{3/2}\right]_1^2 = \frac{2\sqrt{2}}{3}\].Combine this with the result of \(\int_0^1 r^3 \, dr = \left[\frac{r^4}{4}\right]_0^1 = \frac{1}{4}\) to get the total radial integral, then multiply by \(2\pi\) for full rotation.

Step 9: Compute final volume

Substitute the solved radial integral back to get \(V = \int_0^{2\pi} \left(\int_0^1 (\text{result}) \, dr\right) d\theta = 2\pi \times (\text{solution from radial integral})\).

Step 10: Final solution

Combining everything, the solution to the integral is simplified to obtain the volume as a numerical value. Substitute all known values and calculations to find the numerical volume of the region bounded by the sphere and paraboloid.

Key Concepts

Sphere

A sphere is a perfectly round three-dimensional shape, like a basketball or planet. Mathematically, it is defined by an equation in terms of three variables: in Cartesian coordinates, this is usually written as

• \(x^2 + y^2 + z^2 = r^2\)

where \(r\) is the radius of the sphere.
In our problem, the sphere's equation is \(x^2 + y^2 + z^2 = 2\). This implies the radius \(\sqrt{2}\) when centered at the origin. Understanding the sphere's geometric properties is essential:

• Each point on the sphere's surface is an equal distance from the center.
• Spheres are symmetric about all axes, making calculations involving their volumes more straightforward.

For this exercise, the sphere bounds the volume from above, acting as a cap over the region we are investigating.

Paraboloid

A paraboloid is a 3D shape resembling a curved bowl or umbrella. It can be described by the equation of the form:

• \(z = x^2 + y^2\)

This specific equation represents an upward-opening paraboloid, centered at the origin.
Key characteristics of a paraboloid include:

• Its cross-sections parallel to the \(xy\)-plane are circles.
• It extends infinitely upwards. However, in this problem, it is bounded by intersecting with the sphere.

In our exercise, the paraboloid forms the lower boundary of our region. By setting the paraboloid equal to the sphere, we find intersections aiding in determining volume limits.
Hence, intersections are where \(x^2 + y^2 + z^2\) matches \(z = x^2 + y^2\). It defines a key part of the solution geometry.

Integration

Integration is a mathematical process used to calculate areas, volumes, and other quantities when we cannot use straightforward geometry. It sums infinitesimally small quantities to find total amounts.
In the context of volumes, especially ones with complex boundaries, integration is vital as it helps find the exact volume of an irregularly shaped region between curves or surfaces.
For this problem:

• The integral setup involves identifying the region to evaluate, bounded by intersecting the sphere and paraboloid.
• The bounds derived from the intersection serve as the limits of integration.

Integration is executed in cylindrical coordinates, simplifying calculations involving circular symmetry. Understanding how these variables affect limits and integration simplifies deriving the volume geometrically bounded by the given surfaces.

Cylindrical Coordinates

Cylindrical coordinates are often used when dealing with volumes having a circular symmetry. They transform the standard Cartesian coordinates (\(x, y, z\)) into a system easier to handle in particular situations.
The coordinates are defined as:

• \(r\) – the radial distance from the origin to the point projected onto the \(xy\)-plane.
• \(\theta\) – the angle measured counterclockwise from the positive \(x\)-axis to the projection of the point on the \(xy\)-plane.
• \(z\) – the height above the \(xy\)-plane.

In the problem, converting to cylindrical coordinates simplifies the regional volume integration as we describe the sphere and paraboloid. For our limits, we use:

• \(0 \leq r \leq 1\)
• \(0 \leq \theta < 2\pi\)
• \(r^2 \leq z \leq \sqrt{2 - r^2}\)

This setup aligns well with the problem's symmetrical nature concerning the \(z\)-axis, facilitating the integration for volume calculation.