Let R be the region bounded by the curve x=9y-y^2 and the y- axis. Find the volume of the solid resulting from revolving R about the line y= -6.
I believe the integral limits are from y=0 to y-9
i set up
h(x) = 9y-y^2
r(x) = y+6
I am not sure if this is correct
Can anyone help me?
Thanks
I answered my question.
I had made an arithmetic error.
the shape you will get is half a torus.
A torus is a donut or bagel shaped solid.
We have cut ours in half (to make a tasty cream-cheese bagel)
the y-intercepts:
0 = y(9-y)
so y = 0 or y = 9 , You had that
so we need the radius of the outer and the inner rings:
y^2 - 9y = -x
y^2 - 9y + 20.25 = 20.25 -x , I completed the square
(y - 4.5)^2 = 20.25-x
y - 4.5 = ± √(20.25-x)
y = 4.5 ± √(20.25-x)
so the outer radius = 6 + 4.5 + √(20.25-x)
= 10.5 + (20.25-x)^.5
and the inner radius = 6 + 4.5 - √(20.25-x)
= 10.5 - (20.25-x)^.5
volume = π∫( (10.5 + (20.25-x)^.5 )^2 - ((10.5 - (20.25-x)^.5 )^2 dy from y = 0 to 9
= π∫......
I hope I did not mess up the algebra here, you might want to do this yourself. It is rather messy to type
= π∫(42(20.25-x)^(1/2) dy from 0 to 9
= π [-28(20.25-x)^(3/2) ] from 0 to 9
= π( -28(20.25-9)^(3/2) -(-28(20.25-0)^(3/2) )
= π ( - 1056.542... + 2551.5)
= π(1494.9579)
= appr 4696.55
You better check my arithmetic in that last part, so easy to make an error
To find the volume of the solid resulting from revolving the region R about the line y = -6, you can use the method of cylindrical shells.
First, let's rewrite the equation of the curve x = 9y - y^2 in terms of y to make it easier to work with.
x = 9y - y^2
x = y(9 - y)
Now, let's visualize the region R by graphing the curve x = y(9 - y) and the y-axis. This will help us understand the limits of integration.
Next, let's consider a small strip or element in the region R, with a width dy. When this strip is revolved about the line y = -6, it generates a cylindrical shell with radius r and height h.
The radius of the cylindrical shell, r, is given by the distance between the line y = -6 and the curve x = y(9 - y). Since the curve x = y(9 - y) is rotated about y = -6, the radius is given by:
r = x + 6 = (y(9 - y)) + 6
The height of the cylindrical shell, h, is given by the differential dy.
The volume of a cylindrical shell is given by the formula V = 2πrh * dy. Therefore, the volume of the entire solid is the integral of this expression over the limits of integration.
V = ∫[a,b] 2πrh dy
To determine the limits of integration, we need to find the y-values where the curve x = y(9 - y) intersects the y-axis.
Setting x = y(9 - y) equal to zero, we have:
0 = y(9 - y)
This equation is true when y = 0 or when 9 - y = 0. So, the curve intersects the y-axis at y = 0 and y = 9.
Therefore, the limits of integration are from y = 0 to y = 9.
We can now set up the integral to calculate the volume:
V = ∫[0,9] 2π(rh) dy
V = ∫[0,9] 2π((y(9 - y)) + 6)h dy
Evaluating this integral will give you the volume of the solid resulting from revolving R about the line y = -6.