1) A region is bounded by the line y = x and the parabola y = x2 - 6x + 10. What is the volume of the solid generated by revolving the region about the x-axis?
First define the x limits of the enclosed region. They are between the two roots of
x^2 -6x +10 = x
x^2 -7x +10 = 0
(x - 5)(x -2)= 0
You want to integrate between x = 2 and 5.
The function that you integrate is
f(x) = pi*{-[x^2 -6x +10]^2 + x^2} dx.
The y = x curve lies above the parabola in the interval
To find the volume of the solid generated by revolving the region about the x-axis, we can use the method of cylindrical shells.
First, we need to determine the bounds for integration. To find the points of intersection between the line y = x and the parabola y = x^2 - 6x + 10, we set them equal to each other:
x = x^2 - 6x + 10
Rearranging the equation, we get:
x^2 - 7x + 10 = 0
Factoring the quadratic equation, we have:
(x - 2)(x - 5) = 0
Thus, the points of intersection are x = 2 and x = 5.
Now, let's integrate the volume using cylindrical shells. The height of each shell will be the difference between the two functions: (x^2 - 6x + 10) - x.
Therefore, the height of each shell is h(x) = x^2 - 7x + 10.
The radius of each shell is r(x) = x (distance from the x-axis to the parabola).
The infinitesimal volume of each shell is given by:
dV = 2πrh(x) dx
Integrating from x = 2 to x = 5, we get:
V = ∫ [2πx (x^2 - 7x + 10)] dx
Expanding and integrating, we have:
V = 2π ∫ [(x^3 - 7x^2 + 10x) dx]
Using the power rule of integration, we find:
V = 2π [ (1/4)x^4 - (7/3)x^3 + (5/2)x^2 ] evaluated from x = 2 to x = 5
Evaluating the integral, we get:
V = 2π [ (1/4)(5^4) - (7/3)(5^3) + (5/2)(5^2) ] - 2π [ (1/4)(2^4) - (7/3)(2^3) + (5/2)(2^2) ]
V = 2π [ (625/4) - (875/3) + 50 ] - 2π [ (16/4) - (56/3) + 10 ]
V = 2π [ (625/4) - (875/3) + 50 - 4 + (56/3) - 10 ]
V = 2π [ (625/4) - (875/3) - (5/1) + (8/3) ]
V = 2π [ (625/4) - (875/3) - (15/3) + (8/3) ]
V = 2π [ (625/4) - (875/3) - (7/3) ]
V = 2π [ (625/4) - (875/3) - (7/3) ]
V ≈ 2π [ 46.46 ]
V ≈ 291.44π
Therefore, the volume of the solid generated by revolving the region about the x-axis is approximately 291.44π cubic units.
To find the volume of the solid generated by revolving the region about the x-axis, we can use the method of cylindrical shells.
First, let's graph the two functions, y = x and y = x^2 - 6x + 10, to visualize the region bounded by them.
The line y = x is a straight line passing through the origin with a slope of 1, while the parabola y = x^2 - 6x + 10 opens upwards.
To find the points where the two functions intersect, we set them equal to each other:
x = x^2 - 6x + 10
Rearranging this equation, we get:
x^2 - 7x + 10 = 0
Solving this quadratic equation, we find two distinct x-values as the solutions: x = 2 and x = 5.
Now, let's set up the integral to calculate the volume using the cylindrical shell method. The volume of a shell is given by the formula:
V = 2π ∫ [y * (radius)] dx
In this case, the radius is simply the y-value, and y ranges from x to the upper curve (parabola) y = x^2 - 6x + 10, and x ranges from 2 to 5.
The integral to calculate the volume becomes:
V = 2π ∫ [x * (x^2 - 6x + 10)] dx
Integrating this expression, we get:
V = 2π ∫ [x^3 - 6x^2 + 10x] dx
Evaluating this integral from x = 2 to x = 5 will give us the volume of the solid generated by revolving the region about the x-axis.