Asked by Galagan
Hi, can someone confirm if i got the right answer to this question.
SO the region between curves y=x^2+1, y=2x^2-2 and the two axes, Is rotated around the Y-AXIS to form a solid glass vase. Determine the volume of the vase. I got 4*pi. Need confirmation if this is correct, if not then can someone show working out please.
Thank-you
SO the region between curves y=x^2+1, y=2x^2-2 and the two axes, Is rotated around the Y-AXIS to form a solid glass vase. Determine the volume of the vase. I got 4*pi. Need confirmation if this is correct, if not then can someone show working out please.
Thank-you
Answers
Answered by
Steve
The curves intersect at (±√3,4)
So, using the area in the first quadrant, we have, using shells,
v = ∫[0,√3] 2πrh dx
where r=x and h=(x^2+1)-(2x^2-2)=(3-x^2)
v = 2π∫[0,√3] x(3-x^2) dx
= 9π/2
Looks more like a bowl than a vase...
So, using the area in the first quadrant, we have, using shells,
v = ∫[0,√3] 2πrh dx
where r=x and h=(x^2+1)-(2x^2-2)=(3-x^2)
v = 2π∫[0,√3] x(3-x^2) dx
= 9π/2
Looks more like a bowl than a vase...
Answered by
Galagan
Why did you use shells?... it doesn't have a whole in it?
Answered by
Steve
if you use discs, they have holes. Shells are just like nested cylinders, so there are no holes. Using discs (washers), the calculation is more complex, since the lower portion is bounded by the y-axis (solid discs), and the upper portion is bounded by the two curves (washers with holes).
v = ∫[-2,1] πr^2 dy + ∫[1,4] π(R^2-r^2) dy
v = ∫[-2,1] π(y+2)/2 dy + ∫[1,4] π((y+2)/2 - (y-1)) dy
= 9π/4 + 9π/4
= 9π/2
v = ∫[-2,1] πr^2 dy + ∫[1,4] π(R^2-r^2) dy
v = ∫[-2,1] π(y+2)/2 dy + ∫[1,4] π((y+2)/2 - (y-1)) dy
= 9π/4 + 9π/4
= 9π/2
Answered by
Alex
hhmmmm... i used the equation v=integral of pi*y^2 from a to b.
Answered by
Alex
and this got me to Galagan's answer of 4*pi... but still not sure so don't take my answer for it Galagan.
Answered by
Steve
pi y^2 is used when rotating around the x-axis.
Answered by
Galagan
I think you can use pi*y^2 for y-axis too... the cyclindrical shells method however i think is only used when a hole is formed when rotating around one of the axis.
Answered by
Steve
the volume of a disc of thickness k is
pi r^2 k
If the radius is y, it means you are rotating around the x-axis, so the volume of each disc is pi y^2 dx
If rotating around the y-axis, each disc's radius is x, so the volume is pi x^2 dy
Try drawing a diagram, guys. As you saw above, either discs or shells can be used in any situation. Just sometimes one is easier to calculate.
pi r^2 k
If the radius is y, it means you are rotating around the x-axis, so the volume of each disc is pi y^2 dx
If rotating around the y-axis, each disc's radius is x, so the volume is pi x^2 dy
Try drawing a diagram, guys. As you saw above, either discs or shells can be used in any situation. Just sometimes one is easier to calculate.
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