Find the volume when the plane area bounded by y=-x^2-3x+6 and x+y-3=0 is revolved about (1) x=3 and (2) about y=o
B. Find the volu6 of focus generated by revolving the circule x^2+y^2=4 about the line x=3
A: The curves intersect at (-3,6) and (1,2)
So, consider the volume as a set of nested cylinders of thickness dx. Then
v = ∫[-3,1] 2πrh dx
where r=3-x and h=(-x^2-3x-6)-(3-x)
v = ∫[-3,1] 2π(3-x)((-x^2-3x-6)-(3-x)) dx
= 2π∫[-3,1] (3-x)(-x^2-2x+3) dx
= 256π/3
B. Google volume of torus
By integration
To find the volume when a plane area is revolved about a given axis, we can use the method of cylindrical shells or the method of disc/washer method, depending on the shape of the region.
(1) Revolving the plane area bounded by y=-x^2-3x+6 and x+y-3=0 about the line x=3:
- First, let's find the points of intersection between the two curves. We set y=-x^2-3x+6 equal to x+y-3.
-x^2-3x+6 = x+y-3
Let's rearrange this equation:
-x^2-3x+6-x-y+3 = 0
Combine like terms:
-x^2-4x-y+9 = 0
Move all terms to one side:
-x^2-4x-y+9 = 0
Now, let's solve this equation for the values of x and y.
Since we are revolving the area about the line x=3, we want to find the bounds of integration in terms of (x-3).
Let's rewrite y = -x^2-3x+6 as y = -(x+2)(x-3).
Now, the bounds of integration for x would be the x-values of the points of intersection between the curves. So, we need to solve:
-(x+2)(x-3) = x+y-3
Expand:
-x^2 + 3x + 2x - 6 = x+y-3
Combine like terms:
-x^2 + 5x - 6 = x+y-3
Move all terms to one side:
-x^2 + 5x - x + 6 - y - 3 = 0
Simplify:
-x^2 + 4x + 3 - y = 0
Now, square both sides of the equation:
(x-3)^2 = (x^2 - 4x + 3 - y)
Rearrange:
x^2 - 6x + 9 = x^2 - 4x + 3 - y
Simplify:
-6x + 9 = -4x + 3 - y
Move all terms to one side:
-6x + 4x + y = -6
Simplify further:
-2x + y = -6
Now we have a linear equation in terms of (x-3), which gives us the bounds of integration. Let's rearrange it to solve for y:
y = 2x - 6
Now we have our bounds of integration for the volume using the cylindrical shell method: (x-3) from x=1 to x=2, and y from y=2x-6 to y=-(x+2)(x-3).
To calculate the volume using the cylindrical shell method, we use the formula:
V = ∫[a,b]2πx(f(x) - g(x))dx
Where:
- a and b are the bounds of integration (in this case, x=1 and x=2)
- x represents the axis of rotation (in this case, x=3)
- f(x) is the equation of the upper curve (y=2x-6)
- g(x) is the equation of the lower curve (y=-(x+2)(x-3))
Evaluate the integral to find the volume of the region revolved around x=3.
(2) Revolving the circle x^2+y^2=4 about the line x=3:
To find the volume of the shape generated by revolving the circle about x=3, we need to find the equation for the cross-sections created by the revolved shape.
The cross-sections formed will be discs or washers depending on the slice we take.
We need to solve the equation x=3 for y to find the equations for the upper and lower boundaries of the cross-sections.
From x=3, we have:
3^2 + y^2 = 4
9 + y^2 = 4
y^2 = 4 - 9
y^2 = -5
Since the squared term cannot be negative (imaginary), there are no intersections between the circle x^2+y^2=4 and the line x=3.
Therefore, when the circle x^2+y^2=4 is revolved about the line x=3, it does not intersect the line, and thus, no volume is generated.