Problem 25 - Thomas Calculus

Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Sawed-off sphere The lower portion cut from the sphere \(x^{2}+y^{2}+z^{2}=2\) by the cone \(z=\sqrt{x^{2}+y^{2}}\)

Step-by-Step Solution

Answer

The surface area is \(4\pi - 2\pi\sqrt{2}\).

Step 1: Analyze the Surfaces

The sphere is given by the equation \(x^2 + y^2 + z^2 = 2\), and the cone is defined by \(z = \sqrt{x^2 + y^2}\). The task is to find the surface area of the portion of the sphere cut by the cone. The intersection occurs when both equations hold true, which can be set by equating \(z\) in both equations: \(z = \sqrt{x^2 + y^2}\).

Step 2: Establish Parametrization

Recognize that the surface is part of a sphere. Use spherical coordinates for parametrization where \( x = \sqrt{2}\sin \theta \cos \phi \), \( y = \sqrt{2}\sin \theta \sin \phi \), and \( z = \sqrt{2}\cos \theta \). The angle \(\phi\) ranges from \(0\) to \(2\pi\), while the angle \(\theta\) will need to be limited by the condition that corresponds to the intersection with the cone.

Step 3: Identify Bounds of Integration

The intersection at \(z=\sqrt{x^2+y^2}\) in spherical coordinates is \(\sqrt{2}\cos \theta = \sqrt{2}\sin \theta\). Solving \(\cos \theta = \sin \theta\), we get \(\theta = \pi/4\). Therefore, \(\theta\) ranges from \(0\) to \(\pi/4\) to cover the portion above the cone.

Step 4: Surface Area Element

The surface element in spherical coordinates is given by the formula \(dS = R^2 \sin \theta \, d\theta \, d\phi\), where \(R=\sqrt{2}\). Thus, \(dS = 2 \sin \theta \, d\theta \, d\phi\).

Step 5: Set Up the Double Integral

Formulate the double integral to represent the surface area as follows:\[A = \int_{0}^{2\pi}\int_{0}^{\pi/4} 2 \sin \theta \, d\theta \, d\phi\]

Step 6: Evaluate the Integral with Respect to \(\theta\)

Evaluate the integral over \(\theta\):\[\int_{0}^{\pi/4} 2 \sin \theta \, d\theta = 2 [-\cos \theta]_{0}^{\pi/4} = 2 \left(-\cos \frac{\pi}{4} + \cos 0\right) = 2 \left(-\frac{\sqrt{2}}{2} + 1\right) = 2(1 - \frac{\sqrt{2}}{2})\]

Step 7: Evaluate the Integral with Respect to \(\phi\)

Now, evaluate the integral over \(\phi\):\[\int_{0}^{2\pi} (2 - \sqrt{2}) \, d\phi = (2 - \sqrt{2})[\phi]_{0}^{2\pi} = 2\pi(2 - \sqrt{2})\]

Step 8: Compute and Interpret the Result

The result of the integration gives the surface area of the sawed-off sphere:\[A = 4\pi - 2\pi\sqrt{2}\] This is the surface area of the sphere portion above the cone.

Key Concepts

Spherical Coordinates

Spherical coordinates provide a way to describe points in three-dimensional space using three parameters: the radial distance, an azimuthal angle, and a polar angle. They're particularly useful when dealing with problems that have some form of rotational symmetry, like spheres or circular objects.
In the context of this problem, the sphere with equation \(x^2 + y^2 + z^2 = 2\) can be easily described. Instead of using \(x\), \(y\), and \(z\), we use the relations:

• \(x = \, \sqrt{2} \,\sin \theta \,\cos \phi\)
• \(y = \, \sqrt{2} \, \sin \theta \,\sin \phi\)
• \(z = \, \sqrt{2} \,\cos \theta\)

Here, \(\theta\) is the angle from the positive \(z\)-axis, and \(\phi\) is the angle in the \(xy\)-plane from the positive \(x\)-axis. This setup simplifies the mathematics involved in parametrizing the sphere's surface.

Choosing the right range for these angles is essential. For a full sphere, \(\phi\) ranges from \(0\) to \(2\pi\), and \(\theta\) from \(0\) to \(\pi\), but this problem involves a portion of the sphere above a cone, limiting \(\theta\) to \(0\) to \(\frac{\pi}{4}\).

Double Integral

A double integral allows us to calculate a quantity over a surface area in space, by integrating a function over two variables. In our case, the function represents the surface area element. This is crucial when we want to determine the area of a geometrical shape or surface, particularly complex ones like the sawed-off sphere.

When using spherical coordinates for our problem, the double integral becomes the tool to compute the surface area above the cone. The integral is set up as:

• \[\int_{0}^{2\pi}\int_{0}^{\pi/4} 2 \sin \theta \, d\theta \, d\phi\]

The inner integral \(\int_{0}^{\pi/4} 2 \sin \theta \, d\theta\) accounts for the vertical swath of the sphere, covering all angles \(\theta\) from 0 to \(\frac{\pi}{4}\), which are within the region cut by the cone.
The outer integral \(\int_{0}^{2\pi} \, d\phi\) completes a full revolution around the z-axis, ensuring that the surface area is measured in all horizontal directions. Evaluating these integrals sequentially gives the total surface area above the cone.

Surface Area Calculation

To calculate the surface area of complex shapes like portions of a sphere, calculus provides us with integral calculus—specifically using surface integrals over parametric equations. This method effectively adds up infinitesimally small elements across the surface by integrating over the defined limits.

Here, we used the spherical coordinate system, where the surface element \(dS\) is given by \(R^2 \sin \theta \, d\theta \, d\phi\). For our sphere, \(R = \sqrt{2}\) simplifies to a constant multiplier of 2, and simplifies our integral remarkably. Therefore, \(dS = 2 \sin \theta \, d\theta \, d\phi\).

The calculated integral result, \[A = 4\pi - 2\pi\sqrt{2}\], represents the area of the sawed-off sphere above the conical cut. This outcome tells us the exact size of the sphere’s portion that remains above the intersecting cone, using a neat combination of geometric visualization and calculus techniques. Such methods are crucial in solving problems involving curved surfaces, which can often be challenging without the aid of integral calculus.