Volumes of solids of revolutions

In summary, the conversation is discussing the process of finding the volume of a region bounded by the curves y=x^2 and y=x^3 when revolved around the x-axis and y-axis. The first part of the question was solved by subtracting two areas from each other, but the second part was initially attempted using the formula V= π∫ x^2dy, which was incorrect. After discussing using the washer method and substituting x values, it was determined that both methods would result in the same volume.
  • #1
John_Smith01
2
0

Homework Statement



In the question your working with the region bounded by the two curves y=x^2 and y=x^3.

In the first part of the question you had to revolve the region around the x-axis and find the area which I managed to do by subtracting two areas from each other.

The second part of the question asked to revolve the region around the y-axis which I didn't know how to do since I couldn't find (x^3)dy
 
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  • #2
Revolving around the y-axis would give:

V= π∫ x2 dy

so if y=x3, then x = y 1/3 and hence x2 = y2/3
 
  • #3
That's what I tried before but kept getting the answer wrong. I think my textbook is wrong, did you get 1/10pi u^3?
 
  • #4
rock.freak667 said:
Revolving around the y-axis would give:

V= π∫ x2 dy

so if y=x3, then x = y 1/3 and hence x2 = y2/3
Using the washer method, the volume would be: V = π ((x2)2- (x1)2)dy,

where x2=y(1/3) and x1 = y(1/2).
 
  • #5
John_Smith01 said:
That's what I tried before but kept getting the answer wrong. I think my textbook is wrong, did you get 1/10pi u^3?

I do

SammyS said:
Using the washer method, the volume would be: V = π ((x2)2- (x1)2)dy,

where x2=y(1/3) and x1 = y(1/2).

It will end up as the same since OP will have to subtract one from the other while that way will make into one simple integral.
 
  • #6
rock.freak667 said:
...
It will end up as the same since OP will have to subtract one from the other while that way will make into one simple integral.
When OP mentioned "subtracting two areas from each other", I thought he literally meant areas, not volumes. Oh well !
 

Related to Volumes of solids of revolutions

1. What is meant by "volumes of solids of revolutions"?

The term "volumes of solids of revolutions" refers to the calculation of the volume of a 3-dimensional shape created by rotating a 2-dimensional shape around an axis. This is also known as the method of "disk integration" in calculus.

2. What shapes can be used to create volumes of solids of revolutions?

Any 2-dimensional shape can be used to create a volume of solid of revolution, as long as it is rotated around an axis. Some common shapes used include circles, rectangles, and triangles.

3. How is the volume of a solid of revolution calculated?

The volume is calculated by using the formula: V = π ∫ba (f(x))2dx, where π represents pi, a and b are the limits of the integral, and f(x) is the function of the rotated shape.

4. What are some real-life applications of volumes of solids of revolutions?

Volumes of solids of revolutions have many practical applications, such as calculating the volume of a water tank, the volume of a cylindrical container, and the volume of a cone-shaped ice cream cone. They are also used in engineering and architecture to calculate the volume of structures such as bridges, tunnels, and pipes.

5. Are there any limitations to using volumes of solids of revolutions?

While this method is useful for calculating volumes of certain shapes, it does have some limitations. It cannot be used for shapes that have holes or concave regions, and it is not accurate for highly irregular shapes. In addition, it can be a more complex and time-consuming method compared to other volume calculation techniques.

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