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Volume questions (1 Viewer)
- Thread starter richz
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for Ex6.1: 2. (a)
you're rotating the aread between y=4-x<sup>2</sup>, x=2 and y=4 about the x-axis using perpendicular slices.
Each slice = πr<sub>1</sub><sup>2</sup> - πr<sub>2</sub><sup>2</sup>
= π(r<sub>1</sub><sup>2</sup> -r<sub>2</sub><sup>2</sup>)
where r<sub>1</sub> = 4 and r<sub>2</sub> = 4 - x<sup>2</sup>
so you get lim(δx -->0) π∑ (16 - (4-x<sup>2</sup>)<sup>2</sup>)δx (between 0 and 2)
= π∫ 8x<sup>2</sup> - x<sup>4</sup> dx
= π [ 8/3.x<sup>3</sup> - 1/5.x<sup>5</sup>] (between 0 and 2)
= 224π/15
you're rotating the aread between y=4-x<sup>2</sup>, x=2 and y=4 about the x-axis using perpendicular slices.
Each slice = πr<sub>1</sub><sup>2</sup> - πr<sub>2</sub><sup>2</sup>
= π(r<sub>1</sub><sup>2</sup> -r<sub>2</sub><sup>2</sup>)
where r<sub>1</sub> = 4 and r<sub>2</sub> = 4 - x<sup>2</sup>
so you get lim(δx -->0) π∑ (16 - (4-x<sup>2</sup>)<sup>2</sup>)δx (between 0 and 2)
= π∫ 8x<sup>2</sup> - x<sup>4</sup> dx
= π [ 8/3.x<sup>3</sup> - 1/5.x<sup>5</sup>] (between 0 and 2)
= 224π/15
for 2. (c)
You're going to be integrating between 0 and 4 along the y-axis hence you need to express the volume/area of each disc in terms of y.
In this case the radius of each circular slice will be 2 - √(4-y) hence the area of each slice is given by π(2 - √(4-y))<sup>2</sup>δy
Have a go working it from there
.
You're going to be integrating between 0 and 4 along the y-axis hence you need to express the volume/area of each disc in terms of y.
In this case the radius of each circular slice will be 2 - √(4-y) hence the area of each slice is given by π(2 - √(4-y))<sup>2</sup>δy
Have a go working it from there
.yep, thats what i did i got to π(4y-(y^2)/2 + 8/3(4-y)^(3/2) +4y) between 4 and 0KFunk said:for 2. (c)
You're going to be integrating between 0 and 4 along the y-axis hence you need to express the volume/area of each disc in terms of y.
In this case the radius of each circular slice will be 2 - √(4-y) hence the area of each slice is given by π(2 - √(4-y))<sup>2</sup>δy
Have a go working it from there
and i get like 24π where did i go rong??
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You'll hate what went wrongxrtzx said:
. I think you assumed that when you substituted in '0' that the second bracket would simply disappear. Check out what happens when you subtract 8/3(4-y)<sup>3/2</sup> from 24 where y=0 .For 6. you're integrating about the y-axis with perp. strips so you have to talk in terms of y again.
Treat it like a (bi-quadratic?) equation. So y= 6x<sup>2</sup> - x<sup>4</sup>, let u=x<sup>2</sup> ---> y = 6u - u<sup>2</sup>
(*note) the area of each strip will be π(r<sub>1</sub><sup>2</sup> - r<sub>2</sub><sup>2</sup>) = π(x<sub>1</sub><sup>2</sup> - x<sub>2</sub><sup>2</sup>) = π(u<sub>1</sub> - u<sub>2</sub>) where u<sub>1</sub> > u<sub>2</sub>
u<sup>2</sup> - 6u +y = 0
u = 6/2 ± √(36 - 4y)/2
= 3 ± √(9 - y)
since we're integrating between y=0 and y=(9 I think?) then:
3 + √(9 -y) > 3 - √(9 - y)
therefore:
u<sub>1</sub> = 3 + √(9 -y)
& u<sub>2</sub>= 3 - √(9 - y) (since u<sub>1</sub> > u<sub>2</sub>)
See what you can do from there. Let me know if that's not enough (or too much even).
Sorry, that was for question 4. My bad.
Treat it like a (bi-quadratic?) equation. So y= 6x<sup>2</sup> - x<sup>4</sup>, let u=x<sup>2</sup> ---> y = 6u - u<sup>2</sup>
(*note) the area of each strip will be π(r<sub>1</sub><sup>2</sup> - r<sub>2</sub><sup>2</sup>) = π(x<sub>1</sub><sup>2</sup> - x<sub>2</sub><sup>2</sup>) = π(u<sub>1</sub> - u<sub>2</sub>) where u<sub>1</sub> > u<sub>2</sub>
u<sup>2</sup> - 6u +y = 0
u = 6/2 ± √(36 - 4y)/2
= 3 ± √(9 - y)
since we're integrating between y=0 and y=(9 I think?) then:
3 + √(9 -y) > 3 - √(9 - y)
therefore:
u<sub>1</sub> = 3 + √(9 -y)
& u<sub>2</sub>= 3 - √(9 - y) (since u<sub>1</sub> > u<sub>2</sub>)
See what you can do from there. Let me know if that's not enough (or too much even).
Sorry, that was for question 4. My bad.
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This is my working for 6)
δV= π (y-1)^2
y = cosx
δV= π(cosx-1)^2 δx
= π(cos^2 -2cosx+1)
V= π integral (pi/2 -> 0) cos^2x -2cosx +1 dx
= pi/2 integ. (pi/2 -> 0) cos2x +1 -4cosx +2 dx
=pi/2 (sin2x/2 +x - 4sinx +2x) (pi/2 -> 0)
anyone have any ideas why im rong.
δV= π (y-1)^2
y = cosx
δV= π(cosx-1)^2 δx
= π(cos^2 -2cosx+1)
V= π integral (pi/2 -> 0) cos^2x -2cosx +1 dx
= pi/2 integ. (pi/2 -> 0) cos2x +1 -4cosx +2 dx
=pi/2 (sin2x/2 +x - 4sinx +2x) (pi/2 -> 0)
anyone have any ideas why im rong.
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For that one I think you should try the two circle approach again.xrtzx said:This is my working for 6)
δV= π (y-1)^2
y = cosx
δV= π(cosx-1)^2 δx
= π(cos^2 -2cosx+1)
V= π integral (pi/2 -> 0) cos^2x -2cosx +1 dx
= pi/2 integ. (pi/2 -> 0) cos2x +1 -4cosx +2 dx
=pi/2 (sin2x/2 +x - 4sinx +2x) (pi/2 -> 0)
anyone have any ideas why im rong.
δV = π(r<sub>1</sub><sup>2</sup> - r<sub>2</sub><sup>2</sup>)
where r<sub>1</sub> = 1
and r<sub>2</sub> = (y-1)<sup>2</sup> --> δV = π(1 - [cosx -1]<sup>2</sup>)
yielding:
π∫ 2cosx - cos<sup>2</sup>x dx
= π[2sinx - x/2 - 1/4sin2x] between 0 and π/2
= π[2 - π/4]
= 2π - π<sup>2</sup>/4
EDIT: Note, what you did was to find, if you imagine a cylinder, a funnel shape in the cylinder which is actually hollowed out. You were finding area of the hollow part of the solid, so to speak. If you take away the area I found from the area of the greater cylinder i.e π<sup>2</sup>/2 - (2π - π<sup>2</sup>/4) you get your answer ---> 3π<sup>2</sup>/4 - 2π so your method could have been valid had you recognised that you had found the hollow part of the solid and subtracted that from the area of the greater cylinder. That just came to mind so I figured I'd add it to the post. I hope that helps.
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