can anyone help me with the last problem # 14
Find the volume of the given solid generated by revolving the regions bounded by the graphs of the equations about the given lines.
14. y=6-2x-x^2, y=x+6
a. the x-axis
b. the line y=3
The graphs intersect at (-3,3) and (0,6)
Using vertical strips is easiest, since the upper and lower boundaries do not change. That means we integrate on x.
For revolving around the x-axis, that means using discs (washers) of thickness dx, so
v = ∫[-3,0] π(R^2-r^2) dx
where R=6-2x-x^2 and r=x+6
v = ∫[-3,0] π((6-2x-x^2)^2-(x+6)^2) dx = 243π/5
To revolve around y=3, the same idea is used, but each radius is now y-3 instead of y.
To find the volume of the solid generated by revolving the regions bounded by the graphs of the equations about a given line, you can use the disk or washer method. Here's how to solve problem #14 step by step:
1. Start by graphing the equations y = 6 - 2x - x^2 and y = x + 6 to visualize the regions being revolved.
2. To find the points of intersection, set the two equations equal to each other:
6 - 2x - x^2 = x + 6
Simplify the equation:
-2x - x^2 = x
Rearrange to form a quadratic equation:
x^2 + 3x = 0
Factor out an x:
x(x + 3) = 0
Set each factor equal to zero and solve for x:
x = 0 or x = -3
These points, x = 0 and x = -3, represent the points of intersection for the two graphs.
3. Determine which equation is the top curve and which is the bottom curve. You can do this by substituting x = 0 into each equation and comparing the resulting y-values. The equation that gives the larger y-value is the top curve, and the equation that gives the smaller y-value is the bottom curve.
For x = 0:
y = 6 - 2(0) - (0)^2
y = 6
For x = 0:
y = 0 + 6
y = 6
Both equations give the same y-value, so they are both on the top curve.
4. Determine the limits of integration. In this case, we need to find the x-values where the regions start and end. Since the bottom curve is y = 6 - 2x - x^2 and the top curve is y = x + 6, we can find these x-values by setting the two equations equal to each other and solving for x.
6 - 2x - x^2 = x + 6
Simplify the equation:
-2x - x^2 = x
Rearrange to form a quadratic equation:
x^2 + 3x = 0
Factor out an x:
x(x + 3) = 0
Set each factor equal to zero and solve for x:
x = 0 or x = -3
These are the x-values where the regions start and end.
5. Set up the integral based on the method of disks or washers. Since we are revolving the region about different lines (the x-axis and the line y = 3), we need to set up separate integrals for each.
a. Revolving about the x-axis:
The formula for finding the volume using the disk method when revolving about the x-axis is:
V = ∫π(R(x))^2 dx
In this case, the formula for the radius is given by R(x) = y - 0 (since the x-axis is the line y = 0). Substituting the equations into the formula, we have:
V = ∫π((6 - 2x - x^2) - 0)^2 dx
Simplify the equation:
V = ∫π(6 - 2x - x^2)^2 dx
Integrate the equation with the appropriate limits of integration, which are x = -3 to x = 0:
V = ∫π(6 - 2x - x^2)^2 dx
Evaluate the integral from x = -3 to x = 0.
b. Revolving about the line y = 3:
The formula for finding the volume using the washer method when revolving about the line y = 3 is:
V = ∫π(R(x))^2 dx
In this case, the formula for the radius is given by R(x) = (top curve - line of rotation) - (bottom curve - line of rotation), which becomes (x + 6 - 3) - (6 - 2x - x^2 - 3). Substituting the equations into the formula, we have:
V = ∫π((x + 6 - 3) - (6 - 2x - x^2 - 3))^2 dx
Simplify the equation:
V = ∫π(x + 3 - (6 - 2x - x^2))^2 dx
Integrate the equation with the appropriate limits of integration, which are x = -3 to x = 0:
V = ∫π(x + 3 - (6 - 2x - x^2))^2 dx
Evaluate the integral from x = -3 to x = 0.
Solving the integrals will give you the volume of the solid generated by revolving the regions bounded by the graphs of the equations about the given lines.