Introduction: Finding the Volume of an Object Using Integration
Suppose you wanted to find the volume of an object. For many objects this is a very intuitive process; the volume of a cube is equal to the length multiplied by the width multiplied by the height. For a cylinder the volume is equal to the area of the base (radius squared multiplied by π) multiplied by the height of the cylinder.
However, for more elaborate objects, you can't just find the area of the base and multiply by the height. By placing the object along the side of a coordinate plane, you can use integration, a technique I'll cover later, to find the area of a slice of the object. This method will lead into how to find the volume of an object.
In the next step, we'll explore how to find the area contained by a bowl. The same method can be used to find the area of many different objects. As with any math problem, it can take quite a bit of time to learn, but once mastered, can be done in under a minute.
However, for more elaborate objects, you can't just find the area of the base and multiply by the height. By placing the object along the side of a coordinate plane, you can use integration, a technique I'll cover later, to find the area of a slice of the object. This method will lead into how to find the volume of an object.
In the next step, we'll explore how to find the area contained by a bowl. The same method can be used to find the area of many different objects. As with any math problem, it can take quite a bit of time to learn, but once mastered, can be done in under a minute.
Step 1: Finding the Graph of the Bowl
Let's say you have a bowl that is radially symmetrical, and you want to know the volume of the space contained with it.
Imagine that you cut it right down the middle so you have two similar looking pieces of the bowl.
Now we take the flat side of one of these halves and place it against a coordinate plane. By tracing the inside of the bowl, we can see that it is identical to the graph of y=x^2, at least from the points x=0 and x=2. x=0 is where the middle of the bowl would be.
To give an preview of how we'll find the volume of this bowl, we are going to mentally rotate this the part of the graph between x=0 and x=2 around the y-axis using a method known as the Shell Method.
Imagine that you cut it right down the middle so you have two similar looking pieces of the bowl.
Now we take the flat side of one of these halves and place it against a coordinate plane. By tracing the inside of the bowl, we can see that it is identical to the graph of y=x^2, at least from the points x=0 and x=2. x=0 is where the middle of the bowl would be.
To give an preview of how we'll find the volume of this bowl, we are going to mentally rotate this the part of the graph between x=0 and x=2 around the y-axis using a method known as the Shell Method.
Step 2: Infinite Rectangles
To describe how integrating finds the area, imagine a graph like the one shown above.
By drawing a bunch of rectangles that have equal length, finding the area of each individual rectangle and adding them all together would give a close approximation to the actual area of the graph.
Obviously there is a lot of empty space, and this leaves a lot of room for error. The solution here is take make the rectangles smaller, thus reducing the error.
However, no matter how many rectangles we create, there will always be some unaccounted area. So what if we were able to make infinite rectangles?
The idea of infinite rectangles is what is the basic concept behind finding the area with integration. While this could be a whole new subject in itself, the essential idea behind integrating is that we can make each rectangle infinitely small and add each rectangle to one sum to find the total area.
By drawing a bunch of rectangles that have equal length, finding the area of each individual rectangle and adding them all together would give a close approximation to the actual area of the graph.
Obviously there is a lot of empty space, and this leaves a lot of room for error. The solution here is take make the rectangles smaller, thus reducing the error.
However, no matter how many rectangles we create, there will always be some unaccounted area. So what if we were able to make infinite rectangles?
The idea of infinite rectangles is what is the basic concept behind finding the area with integration. While this could be a whole new subject in itself, the essential idea behind integrating is that we can make each rectangle infinitely small and add each rectangle to one sum to find the total area.
Step 3: Setting Up the Integral
With the next two steps I'm going to explain how to integrate. Be aware that in this step, integrating only find the area under the curve, not the volume, which is what we are looking for.
We have the function y=x^2, and we know that we only want the area between x=0 and x=2.
To integrate, we would take our function y=x^2, and put it in an integral as shown above.
The curved line at the beginning simply lets us know that we are integrating. The 0 and 2 means we are integrating the area between x=0 and x=2. The dx at the end means we are integrating with respect to x.
When integrating,simply increase each exponent by 1, and then divide the that whole number by the number you just wrote for the new exponent. In this case, it integrates to ((x^3)/3).
We have the function y=x^2, and we know that we only want the area between x=0 and x=2.
To integrate, we would take our function y=x^2, and put it in an integral as shown above.
The curved line at the beginning simply lets us know that we are integrating. The 0 and 2 means we are integrating the area between x=0 and x=2. The dx at the end means we are integrating with respect to x.
When integrating,simply increase each exponent by 1, and then divide the that whole number by the number you just wrote for the new exponent. In this case, it integrates to ((x^3)/3).
Step 4: Calculating the Area
Now that we integrated the function, we want to find the area between two numbers, x=0 and x=2.
Simply plug in the number on top (x=2) into x. Do the same with the lower number (x=0), and find the difference.
The area, 8/3, is the total area under the function y=x^2 between x=0 and x=2.
Simply plug in the number on top (x=2) into x. Do the same with the lower number (x=0), and find the difference.
The area, 8/3, is the total area under the function y=x^2 between x=0 and x=2.
Step 5: Shell Method
Now that we know how to integrate, we're going to use a method known as the Shell Method. The name Shell Method comes from the fact that you can pick any point on a graph, rotate that point around the y-axis, and it should take on the appearance of a shell. By integrating this rotation, you can find the volume of the function as if it were rotated around the y-axis.
Be aware that integrating finds the volume under the curve, while we actually want to find the volume above the curve.
The Shell Method is found by integrating the radius of an object by the height. The radius of an object represents what point you pick on any point on the graph. Usually the radius is just equal to x. The height is how high the function is at any point on the graph.
After integrating, multiply the number by 2π. This will give us the volume of the function between x=0 and x=2 rotated around the y-axis. The formula is shown above.
Be aware that integrating finds the volume under the curve, while we actually want to find the volume above the curve.
The Shell Method is found by integrating the radius of an object by the height. The radius of an object represents what point you pick on any point on the graph. Usually the radius is just equal to x. The height is how high the function is at any point on the graph.
After integrating, multiply the number by 2π. This will give us the volume of the function between x=0 and x=2 rotated around the y-axis. The formula is shown above.
Step 6: Finding the Radius and Height
In the case of our problem, the radius is x. The height at any point on the graph is x^2.
We can plug these numbers into the Shell Method formula from step 5.
Simplifying x(x^2) equals x^3.
By integrating the with the same method from Steps 3 and 4, we get the number ((x^4)/4).
We can plug these numbers into the Shell Method formula from step 5.
Simplifying x(x^2) equals x^3.
By integrating the with the same method from Steps 3 and 4, we get the number ((x^4)/4).
Step 7: Integrating With the Shell Method
We then plug in around the bounds, x=2 and x=0 and find the difference.
Simplifying this number to get 8π.
This video may help clarify how to integrate using the Shell Method, which is what we covered from Step 5 to Step 7.
Simplifying this number to get 8π.
This video may help clarify how to integrate using the Shell Method, which is what we covered from Step 5 to Step 7.
Step 8: Finding the Area Within the Bowl
If you've been following along, you've probably noticed that we've been finding the area under the curve, while what we want to find is the area above the curve, which is the volume of the bowl.
Since we're looking for the volume within the bowl, simply draw a box around the relevant area and find the volume of the cylinder. Subtracting the volume of the graph from the volume of the cylinder will leave us with the volume of the bowl.
Imagine carving out the the bottom part of our curve from the cylinder, leaving only the top half.
Since we're looking for the volume within the bowl, simply draw a box around the relevant area and find the volume of the cylinder. Subtracting the volume of the graph from the volume of the cylinder will leave us with the volume of the bowl.
Imagine carving out the the bottom part of our curve from the cylinder, leaving only the top half.
Step 9: Finding the Volume of a Cylinder
The volume of a cylinder is calculated by the formula V=π*r^2*h.
The radius is 2 and the height is 4.
Multiplying these numbers together reveals the volume of the cylinder to be 16π.
The radius is 2 and the height is 4.
Multiplying these numbers together reveals the volume of the cylinder to be 16π.
Step 10: Finding the Area Within the Bowl.
Now we have the volume of the entire cylinder and the area outside the curve. By simply finding the difference between the two, we can calculate the volume of the bowl.
16π - 8π = 8π.
The volume in the bowl is 8π, or about 25.1327.
16π - 8π = 8π.
The volume in the bowl is 8π, or about 25.1327.
Step 11: Other Methods
The method we used was known as the Shell Method, since one slice can be rotated around an axis to create a shell.
Other methods that could be used are the Disk Method or the Washer Method.
Using these tools the will allow you to calculate the volume of many different objects. While learning this can take a significant amount time, it should be known that excellence comes with practice, and eventually problems like this can be completed in a matter seconds.
Other methods that could be used are the Disk Method or the Washer Method.
Using these tools the will allow you to calculate the volume of many different objects. While learning this can take a significant amount time, it should be known that excellence comes with practice, and eventually problems like this can be completed in a matter seconds.