Calculating the Volume of a Solid: A Comprehensive Guide

In this article, we will discuss the process of finding the volume of a solid that is bounded by a cylinder. Specifically, we will use polar coordinates and integration to solve this problem. The aim is to provide a detailed explanation of the mathematical steps involved and the importance of using these techniques in understanding the geometry of three-dimensional shapes.

The Problem

The problem at hand is to calculate the volume of the solid bounded by a cylinder. In mathematical terms, the solid is defined by the following equations:

( r 2sin(theta) ) ( r 2z ) ( z 0 )

Here, ( r ) and ( theta ) are the polar coordinates of a point in the plane, and ( z ) represents the vertical axis.

The Solution

To solve this problem, we will use polar coordinates and integrate over the appropriate region.

Step 1: Setting Up the Integral

We start by setting up the triple integral in polar coordinates:

( int_0^{pi} int_0^{2sin(theta)} int_0^{frac{r}{2}} r , dz , dr , dtheta )

This integral represents the volume of the solid. Let's break it down step by step.

Step 2: Integrating with Respect to ( z )

First, we integrate with respect to ( z ):

( int_0^{frac{r}{2}} r , dz r left[ z right]_0^{frac{r}{2}} r left( frac{r}{2} - 0 right) frac{r^2}{2} )

Substituting this result back into the integral, we get:

( int_0^{pi} int_0^{2sin(theta)} frac{r^2}{2} , dr , dtheta )

Step 3: Integrating with Respect to ( r )

Next, we integrate with respect to ( r ):

( int_0^{2sin(theta)} frac{r^2}{2} , dr frac{1}{2} left[ frac{r^3}{3} right]_0^{2sin(theta)} frac{1}{2} left( frac{(2sin(theta))^3}{3} - 0 right) frac{8sin^3(theta)}{6} frac{4sin^3(theta)}{3} )

Substituting this result back into the integral, we get:

( int_0^{pi} frac{4sin^3(theta)}{3} , dtheta )

Step 4: Integrating with Respect to ( theta )

Finally, we integrate with respect to ( theta ). To simplify the integral, we use the substitution ( u cos(theta) ), which implies ( du -sin(theta) , dtheta ) or ( dtheta -frac{1}{sin(theta)} , du ).

When ( theta 0 ), ( u 1 ), and when ( theta pi ), ( u -1 ). The integral becomes:

( int_1^{-1} frac{4}{3} (1 - u^2) (-sin(theta)) , du )

Since ( sin(theta) sqrt{1 - u^2} ), the integral becomes:

( int_{-1}^{1} frac{4}{3} (1 - u^2) , du )

Evaluating the integral, we get:

( frac{4}{3} left[ u - frac{u^3}{3} right]_{-1}^1 frac{4}{3} left( left( 1 - frac{1}{3} right) - left( -1 frac{1}{3} right) right) frac{4}{3} left( frac{2}{3} frac{2}{3} right) frac{4}{3} cdot frac{4}{3} frac{16}{9} )

Therefore, the volume of the solid is ( frac{16}{9} ).

Conclusion

In conclusion, we have successfully calculated the volume of the solid bounded by the given equations using polar coordinates and integration. This method provides a powerful tool for solving similar problems in three-dimensional geometry. Understanding these techniques is crucial for anyone working in fields that require an understanding of spatial relationships, such as engineering, physics, and computer science.

Related Keywords

Solid Volume, Polar Coordinates, Integration