Another volume of revolution around x axis

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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
 

Answers and Replies

  • #2
SammyS
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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
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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
SammyS
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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
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k so the integral of 4ax would be (4/2)ax^2 right?
 
  • #6
SammyS
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When you write
integral of pi R2 dh ,​
how is R related to y ?
 
  • #7
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R^2 and Y^2 end up being the same thing right?
 
  • #8
SammyS
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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
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x=0 to x=a

but I forget how to integrate a constant (a)
 
  • #10
SammyS
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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
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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
SammyS
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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
HallsofIvy
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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?
 
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