Another volume of revolution around x axis

In summary, the author is trying to find the volume of an area bounded by y^2 = 4ax and rotated around the x-axis. They ask for help with the x=a part of the boundary and state that they don't know how to integrate a constant.
  • #1
togo
106
0

Homework Statement


Find the volume of the area bounded by
y^2 = 4ax
and
x=a

rotated around the x-axis.

Homework Equations


integral of pi R^2 dh

The Attempt at a Solution


I just don't know how to handle the x=a part of the boundary. Any hints? thanks
 
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  • #2
togo said:

Homework Statement


Find the volume of the area bounded by
y^2 = 4ax
and
x=a

rotated around the x-axis.

Homework Equations


integral of pi R^2 dh

The Attempt at a Solution


I just don't know how to handle the x=a part of the boundary. Any hints? thanks
Integrate from x=0 to x=a .

The trickier part may be understanding how to handle the integrand .
 
  • #3
well I read lots of books on it and do lots of questions the only way to understand them sometimes is to try it and then see someone do it step by step
 
  • #4
togo said:
well I read lots of books on it and do lots of questions the only way to understand them sometimes is to try it and then see someone do it step by step
Graphing y2 = 4ax will help for this question.
 
  • #5
k so the integral of 4ax would be (4/2)ax^2 right?
 
  • #6
When you write
integral of pi R2 dh ,​
how is R related to y ?
 
  • #7
R^2 and Y^2 end up being the same thing right?
 
  • #8
togo said:
k so the integral of 4ax would be (4/2)ax^2 right?

(I misread this post.)

Yes, that is the correct integral.

What limits of integration should you use?
 
  • #9
x=0 to x=a

but I forget how to integrate a constant (a)
 
  • #10
togo said:
x=0 to x=a

but I forget how to integrate a constant (a)

[itex]\displaystyle \int\,a\,f(x)\,dx=a\,\int f(x)\,dx[/itex]
 
  • #11
I guess that I am not sure how to do that.

for example, ∫(2)dx = 2x + C

but how does that apply here? the answer doesn't have any + in it.
 
  • #12
togo said:
I guess that I am not sure how to do that.

for example, ∫(2)dx = 2x + C

but how does that apply here? the answer doesn't have any + in it.
For a definite integral, the constant of integration has no effect, so it can be ignored.

[itex]\displaystyle \int_{b}^{c}\,a\,f(x)\,dx=a\,\int_{b}^{c} f(x)\,dx[/itex]

In your case b=0 and c=a.

The fact that upper limit of integration is the same as the multiplicative constant is merely a coincidence.

[itex]\displaystyle \int_{0}^{a}\,4a\,x\,dx=4a\,\int_{0}^{a} x\,dx[/itex]
 
  • #13
Typically, one of the very first things you learn about integrals is that the anti-derivative of [itex]x^n[/itex], for n not equal to -1, is [itex](1/(n+1))x^{n+1}+ C[/itex]. What is that for x= 0?
 

Related to Another volume of revolution around x axis

1. What is "another volume of revolution around x axis"?

"Another volume of revolution around x axis" is a mathematical concept in which a 2-dimensional shape is rotated around the x axis to create a 3-dimensional shape.

2. How is this concept used in science?

This concept is used in many areas of science, such as physics, engineering, and geometry. It is often used to calculate the volume of objects with a rotational symmetry, such as cylinders or cones.

3. What is the formula for calculating the volume of revolution around x axis?

The formula for calculating the volume of revolution around x axis is V = ∫π(r(x))^2 dx, where r(x) is the radius of the shape at a given point along the x axis.

4. Are there any real-life examples of this concept?

Yes, there are many real-life examples of this concept. For instance, a water bottle can be seen as a volume of revolution around the x axis, as it is created by rotating a 2-dimensional circle around the x axis. Another example is a Ferris wheel, which can be seen as a volume of revolution of a 2-dimensional shape around the x axis.

5. How does the volume change when the shape is rotated around the x axis?

The volume increases as the shape is rotated around the x axis. This is because the rotation adds more layers to the shape, increasing its volume. The volume will also change depending on the shape of the 2-dimensional figure being rotated and the extent of the rotation.

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