The curves intersect where
-y^2/2 = 2-y^2
That is, at (-2,2)
So, we have the left boundary is x = -y^2/2
and the right boundary is x = 2-y^2
The vertices of the region are at (-2,2),(0,0),(2,0)
So, using shells,
v = ∫2πrh dy
where r = y and h = (2-y^2)-(-y^2/2)=2-y^2/2
v = 2π∫[0,2] y(2-y^2/2) dy
v = 4π
using washers, we have to break the region in two, because of that flat boundary along the x-axis from (0,0) to (2,0)
v = π∫[-2,0]((2-x)-(-2x)) dx + π∫[0,2] (2-x) dx
v = 2π + 2π = 4π
Find the volume of the solid obtained by rotating the region bounded by the curves:
2x = -y^2
y^2 = -x + 2
y = 0
for y >= 0
about the x-axis.
Not sure how to do this one, could anybody help me identify what this looks like and how to identify the outer/inner radius?
Thanks
4 answers
What about the constraint of y = 0
wouldn't that restrict the volume to the part on the left of the y-axis ?
and using washers, we would get only
v = π∫[-2,0]((2-x)-(-2x)) dx
= 2π
wouldn't that restrict the volume to the part on the left of the y-axis ?
and using washers, we would get only
v = π∫[-2,0]((2-x)-(-2x)) dx
= 2π
y=0 is part of the boundary.
y>=0 restricts us to the area above the x-axis.
y>=0 restricts us to the area above the x-axis.
Of course !
For some reason I was thinking x = 0.
Ignore my silly comment , ( I will blame it on "old-timers syndrome" )
For some reason I was thinking x = 0.
Ignore my silly comment , ( I will blame it on "old-timers syndrome" )