Write the integral in one variable to find the volume of the solid obtained by rotating the first‐quadrant region bounded by y = 0.5x2 and y = x 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.
First, let's sketch the region to get a better understanding.
The region bounded by y = 0.5x^2 and y = x in the first quadrant looks like a half parabolic curve and a straight line.
To set up the integral, we need to find the limits of integration and the height of the cylindrical shells.
The region starts at x = 0 and ends at the point where the curves intersect. To find the intersection point, we set the equations equal to each other:
0.5x^2 = x
0.5x^2 - x = 0
x(0.5x - 1) = 0
x = 0, x = 2
So our limits of integration for x are 0 to 2.
The height of each shell is given by the difference between the two curves: h(x) = x - 0.5x^2.
Now, we need to find the radius of each shell. Since we are rotating the region about the line x = 5, the distance from the line x = 5 to any point on the curve is 5 - x.
The volume of each shell can be calculated using the formula for the volume of a cylindrical shell:
dV = 2πrh(x)dx
Integrating this expression from x = 0 to x = 2 will give us the total volume:
V = ∫(0 to 2) 2πr(x)h(x)dx
Substituting the expressions for r(x) and h(x):
V = ∫(0 to 2) 2π(5 - x)(x - 0.5x^2)dx
Simplifying:
V = 2π∫(0 to 2) (5 - x)(x - 0.5x^2)dx
You can evaluate this integral to find the volume of the solid obtained by rotating the given region 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 first step is to graph the two curves in the first quadrant to understand the region being rotated:
Since both curves are symmetric with respect to the y-axis, we can focus on the region where x >= 0.
To calculate the volume, we need to consider an infinitesimally thin vertical strip that is parallel to the y-axis. This strip will rotate around the line x = 5, forming a cylinder.
Let's denote the width of this strip as Δy. We need to find the radius and height of each corresponding cylinder to determine its volume.
To find the radius of each cylinder, we need to consider the difference between the x-coordinate of the line x = 5 and the x-coordinate of the curve y = x at a given y-value. Since the line x = 5 is perpendicular to the y-axis, the radius of each cylinder will be 5 - x.
To find the height of each cylinder, we need to consider the difference in y-values between the curves y = 0.5x^2 and y = x at a given y-value. The height of each cylinder will be x - 0.5x^2.
Now we can set up the integral to sum up the volumes of all these infinitesimally thin cylinders:
V = ∫[y=0 to y=1] 2π(5 - x)(x - 0.5x^2) Δy
To evaluate the integral, we first need to express x in terms of y by solving the equations y = 0.5x^2 and y = x simultaneously. This will give us the interval of integration for y.
For y = 0.5x^2 and y = x, we can set them equal to each other:
0.5x^2 = x
0.5x^2 - x = 0
x(0.5x - 1) = 0
So, we have x = 0 and x = 2 as the bounds of integration for x in terms of y.
Now we can rewrite the integral as:
V = ∫[y=0 to y=1] 2π(5 - x)(x - 0.5x^2) dy
Evaluate this integral to find the volume of the desired solid.
Thank you!
The curves intersect at (0,0) and (2,2)
So, using shells of thickness dx,
v = ∫[0,2] 2πrh dx
where r=5-x and h=x-x^2/2
v = ∫[0,2] 2π(5-x)(x-x^2/2) dx = 16π/3
using discs (washers) of thickness dy,
v = ∫[0,2] π(R^2-r^2) dy
where R=5-y and r=5-√(2y)
v = ∫[0,2] π((5-y)^2-(5-√(2y))^2) dy = 16π/3