Easy Solid of Revolution Problem

In summary, the problem is hard but not impossible, and the solution is to use the disc method and approximate the incremental volume by washers.
  • #1
rwisz
2,268
0
It's not that the problem is extremely hard, it's just got me second guessing myself all the time when I happen to think about it. (Problem from a test last Friday)

Homework Statement


Find the volume of the solid formed by revolving the area bounded by [tex]y=x^2, y=4[/tex] about the x-axis.

Homework Equations


Disc method would be easiest since it keeps everything neat and tidy in terms of [tex]dx[/tex]

[tex]V = \pi\[ \int_a^b [r(x)]^2\,dx.\][/tex]

The Attempt at a Solution


My attempt (and I'm pretty sure I'm right on this) was to include the washer-method.

To find the volume of the area bounded by [tex]y=x^2, y=4[/tex] revolved around the x-axis:
The bounds of integration are easy enough to determine, just setting one function equal to the other yields the interval [-2,2]. I decided to work with the interval [0,2] since it would yield half of the volume of the entire figure, then double it (makes more sense to me).

So my process to solve was to find the volume revolved around the x-axis of the area bounded by [tex]y=4, x=2[/tex] which, using the disc method would be:
[tex]V = \pi\int_0^2 [4]^2\,dx. = 32\pi[/tex]

Then subtracting out the volume revolved around the x-axis of the area bounded by [tex]y=x^2, x=2[/tex] which, using the disc method would be:
[tex]V = \pi\int_0^2 [x^2]^2\,dx. = \frac{32\pi}{5}[/tex]

[tex]V= 32\pi - \frac{32\pi}{5} = \frac{128\pi}{5}[/tex]
Thus the entire volume would be [tex]2[\frac{128\pi}{5}] = \frac{256\pi}{5}[/tex]

Or you could combine both integrals and double them to yield the entire volume of the designated problem.

[tex]V = 2*[\pi\int_0^2 [4]^2 - [x^2]^2\,dx.] = \frac{256\pi}{5}[/tex]This is my attempt, yea or nay?
 
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  • #2
Yea (no such word as yeigh).

Also, neigh is what horses do. You meant nay, I'm sure.

Combining both integrals is what you get when you approximate the incremental volume by washers. [itex]\Delta V = \pi * (R^2 - r^2) * \Delta x[/itex],
where R = 4 and r = x2
 
  • #3
My bad on the stupid mistakes, it's a miracle the rest of it came out right.

Thanks for the agreement. My mind's at ease.
 

Related to Easy Solid of Revolution Problem

1. What is a solid of revolution problem?

A solid of revolution problem is a type of mathematical problem that involves rotating a two-dimensional shape around an axis to create a three-dimensional shape. This is typically done to find the volume or surface area of the resulting solid.

2. How do I solve an easy solid of revolution problem?

To solve an easy solid of revolution problem, you will first need to identify the shape being rotated and the axis of rotation. Then, you can use the appropriate formula (such as the disk method or shell method) to calculate the volume or surface area of the solid.

3. What are some common shapes used in solid of revolution problems?

Some common shapes used in solid of revolution problems include circles, rectangles, triangles, and semicircles. These shapes can be rotated around a variety of axes to create different types of solids.

4. Are there any useful tips for solving solid of revolution problems?

One useful tip for solving solid of revolution problems is to carefully label all given information and draw a clear diagram to visualize the problem. It can also be helpful to break the solid into smaller, simpler shapes and use the appropriate formula for each part.

5. How are solid of revolution problems used in real life?

Solid of revolution problems have many practical applications in fields such as engineering, architecture, and physics. For example, they can be used to calculate the volume of a water tank or the surface area of a cylindrical pipe. They can also be used to model the shape of objects in 3D design software.

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