Problem 51 - Advanced Engineering Mathematics

Question

Use triple integrals and cylindrical coordinates. Find the volume of the solid that is bounded by the graphs of the given equations. $$ x^{2}+y^{2}=4, \quad x^{2}+y^{2}+z^{2}=16, z=0 $$

Step-by-Step Solution

Answer

The volume is \(\frac{16\pi}{3}\).

Step 1: Interpret the problem

We need to find the volume of the region that is bounded by the cylinder described by the equation \(x^2 + y^2 = 4\), the sphere \(x^2 + y^2 + z^2 = 16\), and the plane \(z = 0\). In cylindrical coordinates, the cylinder \(x^2 + y^2 = 4\) becomes \(r = 2\), and the sphere \(x^2 + y^2 + z^2 = 16\) becomes \(r^2 + z^2 = 16\). The plane \(z = 0\) means that we are considering the region from the \(xy\)-plane up to the surface of the sphere.

Step 2: Set up cylindrical coordinates

In cylindrical coordinates, where \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\), the integrals are usually expressed as a volume integral \(\int_0^{2\pi} \int_0^R \int_{g1(r)}^{g2(r)} f(r, \theta, z) \, dz \, r \, dr \, d\theta\). Here \(R = 2\) is the radius of the base of the cylinder, and \(g1(r) = 0\) and \(g2(r) = \sqrt{16 - r^2}\) represents the sphere's surface in terms of \(z\).

Step 3: Setup Triple Integral

The volume is defined by the integral \(V = \int_0^{2\pi} \int_0^2 \int_0^{\sqrt{16 - r^2}} r \, dz \, dr \, d\theta\). The limits for \(z\) are from 0 to \(\sqrt{16 - r^2}\), \(r\) goes from 0 to 2, and \(\theta\) goes from 0 to \(2\pi\). The factor \(r\) appears because of the transformation to cylindrical coordinates.

Step 4: Integrate with respect to z

First, compute the integral with respect to \(z\):\[ \int_0^{\sqrt{16 - r^2}} r \, dz = r[z]_0^{\sqrt{16 - r^2}} = r\sqrt{16 - r^2} \]

Step 5: Integrate with respect to r

Insert the result from Step 4 into the \(r\) integral:\[ \int_0^2 r\sqrt{16 - r^2} \, dr \]Perform the substitution \(u = 16 - r^2\), \(du = -2r \, dr\). The limits change from \(r = 0\) to \(u = 16\), and \(r = 2\) to \(u = 12\). The integral becomes:\[ -\frac{1}{2}\int_{16}^{12} \sqrt{u} \, du \]The integral of \(\sqrt{u}\) is \(\frac{2}{3}u^{3/2}\), so:\[ -\frac{1}{2}\left[ \frac{2}{3}u^{3/2} \right]_{16}^{12} = -\frac{1}{3}\left[(12)^{3/2} - (16)^{3/2}\right] \]Evaluated, it results in \(-\frac{1}{3}(24 - 32) = \frac{8}{3}\).

Step 6: Integrate with respect to θ

The integral becomes:\[ \int_0^{2\pi} \frac{8}{3} \, d\theta = \frac{8}{3} \theta \bigg|_0^{2\pi} = \frac{8}{3} (2\pi) = \frac{16\pi}{3} \]

Step 7: Conclusion

The volume of the solid bounded by the cylinder, sphere, and plane is given by the result of our evaluation, \(\frac{16\pi}{3}\).

Key Concepts

Cylindrical Coordinates

When dealing with three-dimensional problems involving surfaces like cylinders or spheres, cylindrical coordinates simplify the process significantly. Unlike Cartesian coordinates (x, y, z), cylindrical coordinates use a radius, angle, and height (\((r, \theta, z)\)) to describe a point in space.

• The expression for \(x\) and \(y\) is in terms of a radius \(r\) and angle \(\theta\): \(x = r \cos \theta\) and \(y = r \sin \theta\).
• The equation \(x^2 + y^2 = r^2\) becomes helpful since many geometric shapes naturally express with radial symmetry.
• For height, \(z\) remains the same in both Cartesian and cylindrical coordinates.

In our problem, the cylinder's equation \(x^2 + y^2 = 4\) turns into \(r = 2\). The conversion into cylindrical coordinates facilitates the handling of the integral boundaries and ensures accurate volume calculation, especially revolving symmetric solids.

Volume Calculation

Calculating the volume using triple integrals involves layering shallow disks or slices within the solid and then summing these infinitesimal volumes. In cylindrical coordinates, triple integrals become easier for specific symmetric shapes.

• Start by setting bounds for \(\theta\), which usually spans an entire circle from \(0\) to \(2\pi\).
• The radial component \(r\) varies from the center of the shape outwards; in the given task, \(r\) ranges from \(0\) to \(2\) because of the cylinder defined by \(r = 2\).
• The height \(z\) changes within the volume, defined by the sphere's upper boundary and the plane \(z = 0\); hence, \(z\) bounds are from 0 to \(\sqrt{16-r^2}\).

The integration proceeds by first integrating with respect to \(z\), determining the height of the segment, then with respect to \(r\) for radial length, and lastly with respect to \(\theta\), encompassing the rotational symmetry—culminating in the solution for the total volume.

Solid Geometry

Solid geometry deals with measuring volumes and other aspects of three-dimensional shapes. In this exercise, the shape defined is a portion of a sphere truncated by both a plane and a cylindrical boundary.

• The sphere has an equation \(x^2 + y^2 + z^2 = 16\), which translates to radial symmetry where \(x^2 + y^2\) equates to the sphere's base circle and \(z\) denotes its height.
• The cylinder's equation \(x^2 + y^2 = 4\) indicates that all points satisfy \(r = 2\), forming a circular boundary.
• The plane \(z = 0\) is the baseline of our volume, effectively slicing the sphere.

Visualizing these sets the stage for integration. Starting from the top, the sphere meets the cylinder, and integrating over \(z\) helps accommodate the varied height formed by the circle's rotation. Together, these equations represent a capped cylindrical segment of the sphere, describing a unique volume that the integral aptly computes.