Problem 30 - Calculus Early Transcendentals

Question

Use the shell method to find the volume of the following solids. The solid formed when a hole of radius 3 is drilled symmetrically through the center of a sphere of radius 6

Step-by-Step Solution

Answer

Answer: The volume of the solid is approximately 39.6608π cubic units.

Step 1: Set up the shell method formula

We'll use the shell method formula to get the volume of the solid: Volume = ∫[2π * r * height * thickness] The radius ranges from 0 to 3 (the radius of the cylindrical hole) and the height of each cylindrical shell will be determined by the Pythagorean theorem. The thickness of each shell is "dr".

Step 2: Determine the height of each shell

Using the Pythagorean theorem, we can write an equation relating the height (h), the radius of the hole (r), and the radius of the sphere (R): h^2 + r^2 = R^2 In this problem, R = 6. So, the equation becomes: h^2 + r^2 = 6^2 Let's solve for h: h^2 = 6^2 - r^2 h = sqrt(6^2 - r^2)

Step 3: Plug the height and the limits of integration into the shell method formula

Using the height that we found in the previous step and the limits of integration, we can now write the integral that represents the volume of the solid: Volume = ∫[2π * r * sqrt(6^2 - r^2) * dr] from r=0 to r=3

Step 4: Calculate the integral

Now, we'll evaluate the integral to find the volume of the solid: Volume = ∫[2π * r * sqrt(6^2 - r^2) * dr] from r=0 to r=3 = (2π) ∫[r * sqrt(36 - r^2) * dr] from r=0 to r=3 To solve this integral, we can make a substitution. Let u = 36 - r^2. Then, du = -2r dr. We can set up the new integral as follows: Volume = (2π) ∫[-u^(1/2)(1/2) * du] from u=36 to u=27 Evaluating the integral: Volume = (2π)[-1/2 ∫ u^(1/2)du] from u=36 to u=27 = (2π)[-1/2 (2/3)(u^(3/2))] from u=36 to u=27 = (2π)[-(1/3)u^(3/2)] from u=36 to u=27 Applying the limits of integration: Volume = (2π)[-(1/3)(27^(3/2)) - (-(1/3)(36^(3/2)))] Volume = (2π)(-(1/3)[27*sqrt(27) - 36*sqrt(36)]) Volume = (2π)(-(1/3)[81sqrt(3) - 216]) Now, multiply the constants: Volume = (2π)(-27sqrt(3) + 72) We can leave the final answer in this form, or we can calculate the numerical value: Volume ≈ 39.6608π So, the volume of the solid is approximately 39.6608π cubic units.

Key Concepts

Volume of Solids

When we think about volumes of solids, we're referring to the amount of three-dimensional space they occupy. Calculating the volume of a solid object involves integrating in a way that accounts for its entire geometric outline.

For solids of revolution and other complex geometries, traditional methods might be cumbersome. That's where techniques like the shell method come in handy. The shell method allows us to apply integral calculus to determine the volume of more intricate solids, such as those formed by drilling a hole through a sphere.

• The shell method calculates volume by integrating "slices" of a solid, typically vertical cylindrical shells.
• These shells are easy to visualize like the layers of an onion, with each layer adding to the total volume.

Using this method, differentiating the solid into these layers, and applying the integral calculus forms a powerful technique to pinpoint a solid's volume.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, especially pertinent in this scenario. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In this exercise, the theorem helps find the height of each cylindrical shell. Here's how:

• The hypotenuse is the radius of the sphere.
• One of the sides is the radius of the cylindrical hole.
• The other side is the height of the cylindrical shell.

By rearranging the theorem, we find:
\[h^2 + r^2 = R^2\]
Thus, solving for the height \( h \) gives:
\[h = \sqrt{R^2 - r^2}\]
This equation helps determine how far each cylindrical shell extends vertically, with the maximum height occurring at the sphere's center.

Integral Calculus

Integral calculus enables us to compute the total size of an object when dealing with continuously varying shapes or concepts. With the shell method, we're essentially summing a series of infinitely thin cylindrical shells to find the volume of the solid.

In our problem, the integral we set up,\[\int_{0}^{3} 2\pi r \sqrt{36 - r^2} \, dr\]represents the sum of all shells from the center out to the edge of the hole. With each shell's height and radius continually varying, integral calculus allows us to systematically sum their contributions into a coherent whole.

• The radius \( r \) varies from 0 to the hole's radius, 3.
• \( \sqrt{36 - r^2} \) provides the height at each shell through the Pythagorean Theorem.

The evaluated integral results in the solid volume, showing the elegance of calculus in painstakingly bringing together an infinite number of small pieces to form a complete structure.

Cylindrical Shells

Cylindrical shells are a mathematical concept used in the shell method to determine volumes. Imagine a collection of nested cylindrical layers within a solid. Each shell contributes to the total volume.

To understand it better, think about the mechanics of peeling an onion. Each peel you remove is like a cylindrical shell. When you've removed all layers, you're left with a structure whose volume you've accounted for completely with shells.

• The volume contribution of each cylindrical shell is determined by its radius, height, and thickness.
• The thickness is essentially an infinitesimally small variation, \( dr \) in calculus terms.

In this problem, the shell thickness \( dr \) compensates by adding up all these small-shell contributions, hence, calculating the volume thoroughly and accurately. This method provides a clear picture of volume accumulation as one delves from the sphere's outermost surface to its drilled center.