The region R is in the first quadrant and bounded by the x-axis, the y axis, and y= 3+2x-x^2. Find the volume of the solid that results when R is revolving about y+1= 0
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To find the volume of the solid that results when region R is revolving about the line y+1=0, we can use the method of cylindrical shells.
First, let's graph the region R and find the intersection points with the y-axis and x-axis.
1. Intersection points with the y-axis:
Setting x=0 in the equation y=3+2x-x^2 gives y=3.
2. Intersection points with the x-axis:
Setting y=0 in the equation y=3+2x-x^2 gives x^2-2x-3=0, which can be factored as (x-3)(x+1)=0. Therefore, x=3 or x=-1.
Now, let's find the equation of the curve y=3+2x-x^2 in terms of y.
Rewriting the equation, we have x^2 - 2x - (y - 3) = 0.
Using the quadratic formula, x = [2 ± √(4 + 4(y - 3))]/2, which simplifies to x = [1 ± √(y - 1)].
So, we have two equations representing the curve:
x = 1 + √(y - 1) - equation (1)
x = 1 - √(y - 1) - equation (2)
Since we are revolving region R about the line y+1=0, which is equivalent to y=-1, we need to rewrite the equations in terms of y+1.
Replacing x with y+1 in equations (1) and (2), we get:
y+1 = 1 + √(y - 1) - equation (3)
y+1 = 1 - √(y - 1) - equation (4)
Simplifying equations (3) and (4), we have:
√(y - 1) = y - 2 - equation (5)
√(y - 1) = -y - 2 - equation (6)
To find the bounds for integration, we need to find the intersection points of equations (5) and (6).
Squaring both sides of equation (5), we get:
y - 1 = (y - 2)^2
Expanding and simplifying, we have:
y - 1 = y^2 - 4y + 4
Rearranging terms, we get:
y^2 - 5y + 5 = 0
Using the quadratic formula, we find the roots as:
y = (5 ± √5)/2
Since we are in the first quadrant, we take the positive value: y = (5 + √5)/2.
Therefore, the bounds for integration with respect to y will be from y = -1 to y = (5 + √5)/2.
Now, to calculate the volume using the method of cylindrical shells, we integrate along the y-axis from the lowest bound to the highest bound:
V = ∫[from -1 to (5 + √5)/2] 2π (y + 1) (x) dy
Since x = 1 + √(y - 1) - equation (1), we substitute it in:
V = ∫[from -1 to (5 + √5)/2] 2π (y + 1) (1 + √(y - 1)) dy
Integrating this expression will give us the volume of the solid obtained by revolving region R about the line y+1=0.