Simple calculus volumes integration

In summary: The volume of the disc shown shaded in your figure is a circular-based cylinder of thickness dx. For the moment, let's forget about the hole in the disc. At any distance x, the circular face of that disc is of radius = y. Using the area of a circle formula, and the thickness of the disc, what is the expression for the volume of just that thin disc shown shaded? (No calculus is involved in answering this.)In summary, the volume of the disc shown shaded in your figure is πr^3.
  • #1
togo
106
0

Homework Statement


Find the volume of this equation, revolved around x axis


Homework Equations


y=x^2
y^2=x

The Attempt at a Solution


1) (pi)(r^2)
2) r = x^2
3) (pi)((x^2)^2)
4) (pi)(x^4)

now to integrate

5) (pi)(1/5(x^5))

since x = 1, and 1^5=1, 1/5=1/5

pi/5?

there are two equations here (y=x^2 & y^2=x), are these two somehow combined for the result

the answer is supposed to be 3pi/10

thanks.
 
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  • #2
You're not looking for the volume of the "equation", you're looking for the volume of the object that's formed when you revolve a 2-d object around the x-axis.

This may help:

http://www.wyzant.com/Help/Math/Calculus/Integration/Finding_Volume.aspx

Also, were those the actual equations given to you?

y = x^2 and y^2 = x?
 
  • #3
yes the question gave me those two equations specifically. The picture of the answer shows two curves.
 
  • #4
3) (pi)((x^2)^2) ?
 
  • #5
((x^2)^2) = x^4?

could someone just do this, I've wracked my brain on it hard enough already.
 
  • #6
Have you sketched the graphs? Are you finding the volume of material used in molding the walls of that 3D object?

Perhaps you should post the solution.
 
  • #7
ea1ij9.jpg


question 7
 
  • #8
But you've cut off that part that was going to answer my question! :cry: :cry:
 
  • #9
lol sorry

szkgug.jpg
 
  • #10
NascentOxygen said:
Have you sketched the graphs? Are you finding the volume of material used in molding the walls of that 3D object?

Perhaps you should post the solution.
Okay, so from the solution we can see that the question does indeed require that you, for example, find the volume of clay needed to make the walls of that aforementioned jar. :cool:

Now that we all understand the question...are you right to finish it?
 
  • #11
obviously not
 
  • #12
It might be clearer if we attack this in two steps:

① Find the volume of the generated solid enclosed within the outer curve (viz., x=y²) for 0≤ x ≥1,

② Find the volume of the generated solid enclosed within the inner curve (viz., y=x²) for that same domain.

Finally, subtract these volumes to determine the difference.

The volume of the disc shown shaded in your figure is a circular-based cylinder of thickness dx. For the moment, let's forget about the hole in the disc. At any distance x, the circular face of that disc is of radius = y. Using the area of a circle formula, and the thickness of the disc, what is the expression for the volume of just that thin disc shown shaded? (No calculus is involved in answering this.)
 
Last edited:

1. What is the concept of integration in simple calculus?

Integration in simple calculus is the process of finding the area under a curve by dividing it into smaller rectangles and summing up their areas. It is the inverse process of differentiation, which is used to find the slope of a curve at a specific point.

2. How is integration used to calculate volumes?

Integration is used to calculate volumes by breaking down a three-dimensional shape into infinitely small slices and summing up the volumes of these slices using the integration formula. This process is known as volume integration and is typically used for finding the volumes of irregular shapes or objects with curved surfaces.

3. What is the difference between definite and indefinite integration?

Definite integration is used to find the exact value of an integral within a specific range, while indefinite integration is used to find the general formula for an integral without specifying its limits. Definite integration results in a numerical value, while indefinite integration results in a formula with a constant of integration.

4. Can integration be used to find the volumes of solids of revolution?

Yes, integration can be used to find the volumes of solids of revolution, which are three-dimensional shapes formed by rotating a two-dimensional shape around an axis. By using the disk or shell method, we can apply integration to calculate the volume of these solids.

5. What are some real-world applications of simple calculus volumes integration?

Simple calculus volumes integration has various real-world applications, such as calculating the volume of irregularly shaped objects in engineering and architecture, determining the amount of fluid in a tank or pipe, and finding the mass and center of mass of an object with varying density. It is also used in physics to calculate the moment of inertia and in economics to estimate consumer and producer surplus.

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