Problem 33 - Calculus Early Transcendentals

Question

Suppose a sphere of radius \(r\) is sliced by two horizontal planes \(h\) units apart (see figure). Show that the surface area of the resulting zone on the sphere is \(2 \pi r h,\) independent of the location of the cutting planes.

Step-by-Step Solution

Answer

Question: Prove that the surface area of a zone on a sphere is given by the formula \(2\pi r h\), where \(r\) is the radius of the sphere, and \(h\) is the distance between the two horizontal planes slicing the sphere. Answer: The surface area of a zone on a sphere can be derived by the surface area of revolution formula. Through a series of steps involving the sphere's equation, Pythagorean theorem, and simplification, we arrive at the final formula: \(A = 2\pi r h\). Thus, the surface area of a zone on a sphere is given by \(2\pi r h\).

Step 1: Visualize the problem and establish coordinate system

We can start by visualizing the sphere with the two cutting planes. Let's assume the sphere is centered at the origin of a 3D Cartesian coordinate system, with radius \(r\). We are given that the distance between two horizontal cutting planes is \(h\). Let's denote the points on the upper cutting plane as \((x_1, y_1)\) and \((x_2, y_2)\) on the lower cutting plane. These points lie on the surface of the sphere.

Step 2: Write the equation of the sphere

Since the sphere is centered at the origin and has radius \(r\), its equation is given by: \(x^2 + y^2 + z^2 = r^2\)

Step 3: Find the height of the points on the cutting planes

Since the two cutting planes are \(h\) units apart, we have: \(y_2 = y_1 - h\) From the equation of the sphere and since \(z = 0\) on the cutting planes, we have: \(x_1^2 + y_1^2 = x_2^2 + y_2^2 = r^2\) Therefore: \(y_1^2 = r^2 - x_1^2\) \(y_2^2 = r^2 - x_2^2\)

Step 4: Find the length of the arc formed by the curve of intersection on the sphere

As we follow an intersection curve from \((x_1, y_1)\) to \((x_2, y_2)\), we form an arc on the sphere's surface. Let's denote the length of this arc as \(s\). Then using the Pythagorean theorem: \(s^2 = (y_1 - y_2)^2 + (x_1 - x_2)^2\) Substitute \(y_2 = y_1 - h\): \(s^2 = h^2 + (x_1 - x_2)^2\) Now, substitute \(y_1^2 = r^2 - x_1^2\) and \(y_2^2 = r^2 - x_2^2\): \(s^2 = (r^2 - x_1^2) - (r^2 - x_2^2) - h^2 + (x_1 - x_2)^2\) Simplify the expression: \(s^2 = x_2^2 - x_1^2 - h^2 + (x_1 - x_2)^2\)

Step 5: Calculate the surface area of the zone

To find the surface area of the zone, we'll apply the surface area of revolution formula for the arc's length \(s\) with respect to the vertical axis and a radius of \(r\): \(A = 2\pi r s\) Substitute the expression for \(s^2\): \(A = 2\pi r \sqrt{x_2^2 - x_1^2 - h^2 + (x_1 - x_2)^2}\) Notice that the term \((x_1 - x_2)^2\) cancels out with the \(x_2^2 - x_1^2\), and we are left with: \(A = 2\pi r \sqrt{-h^2}\) Since we know that the distance between the two planes \(h\) is positive, we can simplify the expression as: \(A = 2\pi r h\) This proves that the surface area of a zone on a sphere is given by the formula \(2\pi r h\), independent of the location of the cutting planes.