Write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5.
To find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5, we can use the method of cylindrical shells and set up an integral.
First, let's visualize the region in the first quadrant bounded by the given curves:
```
y = 0.5x^2
y = x
```
To rotate this region about the line x = 5, we will consider a small strip or "shell" of width Δx at an arbitrary x-coordinate.
The height of the shell, which is the length of the curved surface is given by the difference between the y-values of the two curves at that x-coordinate. So, the height of the shell is:
height = (x - 0.5x^2)
To calculate the circumference or the curved surface area of the shell, we need the distance from the axis of rotation (x = 5) to the shell. Since the axis of rotation is at x = 5, the distance from the axis to the rightmost edge of the shell is:
radius = (5 - x)
Now, the curved surface area of the shell is given by 2π(radius)(height).
Therefore, the volume element (dV) is equal to 2π(radius)(height)(Δx).
To obtain the total volume, we need to integrate the volume element over the range of x-values that span the region. The limits of integration will be the x-values at which the two curves intersect, which can be found by setting them equal to each other:
0.5x^2 = x
Solving this equation, we get x = 0 and x = 2 as the points of intersection.
Now, we can set up the integral to find the volume:
V = ∫[from 0 to 2] 2π(radius)(height) dx
Substituting the expressions for radius and height, we get:
V = ∫[from 0 to 2] 2π(5 - x)((x - 0.5x^2)dx
Now, we can evaluate this integral to find the volume of the solid obtained by rotating the region bounded by the given curves about the line x = 5.
To find the volume of the solid obtained by rotating the first-quadrant region bounded by the curves y = 0.5x^2 and y = x about the line x = 5, we can use the method of cylindrical shells.
The volume of the solid can be found by integrating the differential volumes of the cylindrical shells over the desired region.
First, let's determine the limits of integration for the integral:
The region bounded by the curves y = 0.5x^2 and y = x in the first quadrant is formed between the x-values where the two functions intersect.
Setting the two equations equal to each other, we have:
0.5x^2 = x
Simplifying this equation gives us:
0.5x^2 - x = 0
Multiplying through by 2 so that we have integer coefficients, we get:
x^2 - 2x = 0
Factoring out the common factor of x gives us:
x(x - 2) = 0
So, the x-values where the two functions intersect are x = 0 and x = 2.
Therefore, the limits of integration for the integral will be from x = 0 to x = 2.
Now, let's set up the integral to find the volume:
V = ∫ [2π * radius * height] dx
The radius of each cylindrical shell is the distance from the line x = 5 to the curve y.
The height of each cylindrical shell is the differential length dx.
The radius can be calculated as the difference between the x-value and 5 (distance from x = 5):
radius = (x - 5)
The height is given by the difference between the two functions at a given x-value:
height = (x - 0.5x^2) - x
Simplifying further, we have:
height = (1 - 0.5x^2)
Substituting these values into the integral, we get:
V = ∫ [2π * (x - 5) * (1 - 0.5x^2)] dx
Now, integrate this expression with respect to x over the limits x = 0 to x = 2:
V = ∫[0 to 2] [2π * (x - 5) * (1 - 0.5x^2)] dx
Evaluate this integral to find the volume of the solid obtained by rotating the region about the line x = 5.
using shells,
v = ∫[0,2] 2πrh dx
where r = 5-x and h=x-0.5x^2
or, using discs,
v = ∫[0,2] π(R^2-r^2) dy
where R=5-y and r=5-√(2y)