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
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
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
y=0 is part of the boundary.
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" )
To find the volume of the solid obtained by rotating the given region about the x-axis, we can use the method of cylindrical shells.
First, let's determine what the region bounded by the curves looks like.
The first equation, 2x = -y^2, represents a downward opening parabola, symmetric with respect to the y-axis. This parabola is located in the fourth quadrant since y^2 is always non-negative while x is negative.
The second equation, y^2 = -x + 2, represents an upward opening parabola, also symmetric with respect to the y-axis. This parabola is located in the first quadrant since y^2 is always non-negative while x is positive.
To find the points of intersection between the curves, we set the two equations equal to each other:
2x = -y^2
y^2 = -x + 2
Solving this system of equations, we get:
y^2 = -2x
y^2 = -x + 2
By substituting -2x for x in the second equation, we can find the points of intersection:
y^2 = 4x
y^2 = -x + 2
4x = -x + 2
5x = 2
x = 2/5
Hence, the two curves intersect at x = 2/5.
To determine the inner and outer radii for the cylindrical shells, we need to consider the distance between the curves and the distance from the curves to the x-axis.
From the given equations, it is evident that the parabolas are symmetric with respect to the y-axis. Therefore, the curves on either side of the y-axis will produce symmetrical cylindrical shells.
To represent the region as a solid of rotation, we consider the curves as y = f(x) and y = g(x). In this case, y = 0 is the x-axis and f(x) = -sqrt(-2x) is the outer curve while g(x) = sqrt(4x) is the inner curve.
Since the region is bounded by the x-axis, the lower limit of integration is x = 0. The upper limit of integration is x = 2/5, the x-coordinate where the two curves intersect.
The volume of the solid can be calculated using the following formula:
V = ∫[a, b] 2πx |f(x) - g(x)| dx
where a and b are the limits of integration.
Substituting the values into the formula, we have:
V = ∫[0, 2/5] 2πx |-sqrt(-2x) - sqrt(4x)| dx
Now, you can solve the integral to find the volume of the solid using calculus techniques, such as integration by parts or substitution.
Once you have the definite integral set up, you can evaluate it numerically or symbolically to find the volume.