Consider the region trapped between the graphs of y=9-x^2 and y=11-3x.
a) find the area of this region
b) set up an integral which computes the volume of the solid of rotation obtained by rotating this region about the x-axis
c) set up an integral which computes the volume of the solid of rotation obtained by rotating this region about the y-axis
a) To find the area of the region trapped between the graphs, we need to find the points of intersection of the two curves. We can do this by setting the two equations equal to each other and solving for x.
9 - x^2 = 11 - 3x
Rearranging the equation, we get:
x^2 - 3x + 2 = 0
Factoring the quadratic equation, we have:
(x - 2)(x - 1) = 0
So, the two points of intersection are x = 1 and x = 2.
To find the area of the region, we integrate the difference of the two functions over the interval [1, 2]:
Area = ∫(11 - 3x - (9 - x^2)) dx from 1 to 2
b) To set up an integral to compute the volume of the solid of rotation obtained by rotating this region about the x-axis, we use the method of cylindrical shells.
The radius of each shell is the x-coordinate, and the height of each shell is the difference between the two functions, y = 9 - x^2 and y = 11 - 3x.
The volume of each shell is given by:
dV = 2πrh dx
where r is the radius and h is the height.
Integrating the volume with respect to x from x = 1 to x = 2, the integral becomes:
Volume = ∫(2πx(11 - 3x - (9 - x^2))) dx from 1 to 2
c) To set up an integral to compute the volume of the solid of rotation obtained by rotating this region about the y-axis, we use the method of disks or washers.
We need to express the equations in terms of x since we are rotating about the y-axis. Rearranging the equations, we have:
x = ±√(9 - y)
x = (11 - y)/3
To find the limits of integration, we need to find the y-values at which the two curves intersect. Setting the equations equal to each other, we have:
√(9 - y) = (11 - y)/3
Squaring both sides, we get:
9 - y = (11 - y)^2/9
Simplifying, we have:
81 - 9y = (11 - y)^2
Expanding and rearranging the equation, we get:
y^2 - 22y + 40 = 0
Factoring the quadratic equation, we have:
(y - 2)(y - 20) = 0
So, the two points of intersection are y = 2 and y = 20.
The integral to compute the volume becomes:
Volume = π ∫[(√(9 - y))^2 - ((11 - y)/3)^2] dy from 2 to 20