Volumes Q (1 Viewer)

[shaon0]

y=2sqrt(1-x^2) and y= sqrt(1-x^2) intersect at (1,0) and (-1,0). Find the volume of the enclosed when the region is rotated around:
a) The x-axis
b) The y-axis.

I have done a) but i need help on b).

 

[lolokay]

for b) you have y=2sqrt(1-x^2) and y= sqrt(1-x^2)
so x²=1 - y²/4 and x² = 1 - y²
you're finding the volume between y=0->2 (since 2 is the maximum y value of the 2 functions)

so pi Int 2,0 [3/4 y².dy]
= pi[1/4 y³.dy]2,0
= 2/3pi

sounds about right to me.

(you'll notice that the two graphs are of a semi-circle, with r=1, and semi-ellipse, with height 2, width 2 - which will have double the volume of the semi-circle when rotated
so the volume of area between them will be equal to the volume of the semi circle, which is 2/3 r³ pi, so 2/3 pi for r=1)

 

[shaon0]

for b) you have y=2sqrt(1-x^2) and y= sqrt(1-x^2)
so x²=1 - y²/4 and x² = 1 - y²
you're finding the volume between y=0->2 (since 2 is the maximum y value of the 2 functions)

so pi Int 2,0 [3/4 y².dy]
= pi[1/4 y³.dy]2,0
= 2/3pi

sounds about right to me.

(you'll notice that the two graphs are of a semi-circle, with r=1, and semi-ellipse, with height 2, width 2 - which will have double the volume of the semi-circle when rotated
so the volume of area between them will be equal to the volume of the semi circle, which is 2/3 r³ pi, so 2/3 pi for r=1)

where did you get the 3/4 y^3 from? and the step after that should be 2pi.

 

[lolokay]

ah yeah, that method didn't actually work..

originally what I had was:

pi/4 Int 2,0 [4 - y²] - pi Int 1,0 [1 - y²]
pi/4 (8 - 8/3) - pi + pi/3
= 2pi/3

I had gotten the 3/4 y² from subtracting 1 - y2 from 1 - y2/4

 

[Aerath]

I hope I'm right. =\

 

[shaon0]

I hope I'm right. =\

thanks, but i already have the answer.