Integration Problems

  • Thread starter Sombra
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if I want to find the volume of y= (x + 2) ^(1/2) about the x-axis with the boundaries of -1 and 1, then would it be :

V = pi times integral of (x+2) dx?? Its confusing trying to set it up. The volume of a section would be v = pi r^2 (delta x)

Also I'm having trouble finding the volume of the region enclosed by y = (2-x)(2+x), y = 3x, and the y- axis when rotated around the x - axis. It's hard to even picture what the figure will be like, much harder to find the volume. Please point me in the right direction! Thanks!!
 

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  • #2
NateTG
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Sombra said:
if I want to find the volume of y= (x + 2) ^(1/2) about the x-axis with the boundaries of -1 and 1, then would it be :

V = pi times integral of (x+2) dx?? Its confusing trying to set it up. The volume of a section would be v = pi r^2 (delta x)

Also I'm having trouble finding the volume of the region enclosed by y = (2-x)(2+x), y = 3x, and the y- axis when rotated around the x - axis. It's hard to even picture what the figure will be like, much harder to find the volume. Please point me in the right direction! Thanks!!
You seem to have the first part, but don't forget the limits of integration.
[tex]\pi\int_{-1}^{1}(x+2) dx[/tex]

For the second part, one thing that jumped out at me was:
[tex]y=(2-x)(2+x)=2^2-x^2=4-x^2[/tex]
is a parabola that opens downward, so it will hit the line
[tex]y=3x[/tex] in two (or zero) places.

So, let's look for intersections by solving:
Subtract
[tex]y=3x[/tex]
from
[tex]y=x^2-4[/tex]
to get
[tex]0=x^2-3x-4[/tex]
so we can use the quadratic formula
[tex]x=\frac{3\pm\sqrt{9-4(1)(-4)}}{2}=\frac{3 \pm 5}{2}[/tex]

That tells us that the functions intersect where [itex]x=-1[/itex] and where [itex]x=4[/itex]. Those will be the limits of integration. A quick check, say at [itex]x=0[/itex] confirms that [itex]4-x^2 > 3x[/itex] in the region.

So, to calculate the volume, you can (equivalently) calculate the volume from [tex]y=4-x^2[/tex]
on and then subtract the volume from
[tex]y=3x[/tex]
on the region.
Or you can integrate using annuli (annuluses?) instead of disks.
 
  • #3
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So, for the first one, the answer would be 4pi, correct?

For the second one, I thought the 2 functions interesected at (1,3), so I said the volume was pi times integral (with boundaries of 0 and 1) of [(4-x^2) -(3x)] dx since it has to be subracted. But I think I need to find the integral of r^2 times delta x, so would i have to square (4-x^2) and make it (16-x^4) then subtract 3x (or would it be 9x^2)??? Thanks
 

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