Region R is bounded by x=2-2y^2 and x=1-y^2
find the volume of the solid generated if R is revolved about the y-axis
Anybody help me with this? thanks!
The graphs intersect at (0,-1) and (0,1).
If this area is spun around the y-axis, you can think of the solid as a bunch of thin discs, of thickness dy, each with a hole in the middle.
So, the area is
a = ∫[-1,1] π(R^2-r^2) dy where
R = 2-2y^2 and r = 1-y^2
a = ∫[-1,1] π((2-2y^2)^2-(1-y^2)^2) dy
= 3π∫[-1,1] (y^4-2y^2+1) dy
or, using the symmetry of the figure,
a = 6π∫[0,1] (y^4-2y^2+1) dy
= 6π(1/5 - 2/3 + 1)
= 16π/5
You can also think of the shape as a set of nested shells, each of thickness dx.
x = 2-2y^2, so y = √(1 - x/2)
x = 1-y^2, so y = √(1-x)
We can again exploit the symmetry, but we still have to divide the region into two parts:
for 0<=x<=1, the height of each shell is just the difference in y-values.
For 1<=x<=2, the height is just the y-value, since the x-axis is the lower boundary.
a/2 = ∫[0,1] 2πx(√(1-x/2)-√(1-x)) dx + ∫[1,2] 2πx(√(1-x/2)) dx
= 2π/15 (12-7√2) + 14π/15 √2
= 8π/5
So, a = 16π/5
Great! Except that you wrote "area" instead volume....:P
To find the volume of the solid generated when region R is revolved about the y-axis, you can use the method of cylindrical shells.
First, let's find the limits of integration. To do this, we need to find the y-coordinates of the points where the two curves intersect.
Setting the two equations equal to each other:
2 - 2y^2 = 1 - y^2
Simplifying:
y^2 = 1
Taking the square root:
y = ±1
Since we are revolving around the y-axis, the limits of integration for y will be from -1 to 1.
Now, let's set up the integral for the volume using cylindrical shells. The volume of each shell is given by the formula:
dV = 2πrhdy
Where:
- dV is the volume of the shell
- r is the distance from the y-axis to the curve (in this case, the value of x)
- h is the height of the shell (in this case, the differential change in y)
The equation of the curve x=2-2y^2 can be rewritten as x = 2y^2 - 2.
So, the distance from the y-axis to the curve (r) is given by r = 2y^2 - 2.
The height of the shell (h) is given by h = dy.
Now, we can set up the integral:
V = ∫(from -1 to 1) 2π(2y^2 - 2)dy
Integrating this expression will give you the volume of the solid generated when region R is revolved about the y-axis.