Find the volume of the solid generated by rotating about the y axis the area in the first quadrant bounded by the following curve and lines.
y=x^2, x=0, y=2.
use discs or shells.
v = ∫[0,2] πr^2 dy
where r = x
or
v = ∫[0,√2] 2πrh dx
where r=x and h=y
What do you mean my discs and shells?
To find the volume of the solid generated by rotating the given area about the y-axis, we can use the method of cylindrical shells.
First, let's plot the curve and the lines to get a better understanding of the problem.
The curve y = x^2 is a parabola that opens upward and passes through the points (0, 0) and (1, 1).
The line x = 0 is the y-axis, and the line y = 2 is a horizontal line passing through the point (0, 2).
To find the bounds of integration, we need to determine the values of x that correspond to the curve and the lines intersecting.
To find the intersection point between the curve y = x^2 and the line y = 2, we can set x^2 = 2 and solve for x:
x^2 = 2
x = ±√2
Since we are working in the first quadrant, we can ignore the negative solution. Therefore, the bounds of integration for x are from 0 to √2.
Now we can set up the integral to find the volume.
The volume of a cylindrical shell can be expressed as:
∫[lower bound][upper bound] 2πx * (f(x) - g(x)) dx,
where f(x) is the upper curve and g(x) is the lower curve.
In this case, the upper curve is y = 2 and the lower curve is y = x^2.
So the integral becomes:
V = ∫[0][√2] 2πx * (2 - x^2) dx.
Let's solve this integral step-by-step:
1. V = ∫[0][√2] 2πx * (2 - x^2) dx
2. V = 2π ∫[0][√2] (2x - x^3) dx
3. V = 2π [x^2 - (x^4/4)] |[0][√2]
4. V = 2π [(√2)^2 - ((√2)^4/4)] - 2π [(0)^2 - ((0)^4/4)]
5. V = 2π [2 - 2/4]
6. V = 2π [2 - 1/2]
7. V = 2π [3/2]
8. V = 3π.
Therefore, the volume of the solid generated by rotating the given area about the y-axis is 3π cubic units.
To find the volume of the solid generated by rotating the area bounded by the curve y = x^2, the x-axis, and the lines x = 0 and y = 2 about the y-axis, we can use the method of cylindrical shells.
The first step is to sketch the area in the first quadrant. The curve y = x^2 is a parabola that opens upward and passes through the points (0, 0) and (1, 1). The line x = 0 is the y-axis, and the line y = 2 is a horizontal line passing through the point (0, 2). The area bounded by these curves looks like a quarter-circle shape.
Next, we need to find the bounds of integration. Since we are rotating about the y-axis, the radius of each cylindrical shell will be the x-coordinate of the point on the curve y = x^2. The outer radius of the solid is 2 because it is bounded by the line y = 2. Therefore, the bounds of integration for the cylindrical shells will be from y = 0 to y = 2.
Now, we can express the volume of each cylindrical shell. The height of each cylindrical shell will be the difference between the x-coordinate of the point on the curve y = x^2 and the x-coordinate of the y-axis, which is 0. The differential height can be expressed as dy. The circumference of each cylindrical shell can be approximated as 2π times the radius, which is x.
Thus, the volume of each cylindrical shell is given by dV = 2πx dy.
To find the total volume, we integrate the volume of each cylindrical shell from y = 0 to y = 2:
V = ∫[0,2] 2πx dy
To evaluate this integral, we need to express x in terms of y. Since y = x^2, we can solve for x:
x = √y
Substituting this into the integral, we have:
V = ∫[0,2] 2π(√y) dy
Now, we can evaluate the integral:
V = 2π ∫[0,2] (√y) dy
We can simplify this integral as follows:
V = 2π ∫[0,2] y^(1/2) dy
Using the power rule for integration, which states that ∫ x^n dx = (1/(n+1))x^(n+1) + C, where C is a constant of integration, we integrate as follows:
V = 2π [(2/3)y^(3/2)] |[0,2]
Plugging in the limits of integration, we get:
V = 2π [(2/3)(2)^(3/2) - (2/3)(0)^(3/2)]
Simplifying further, we have:
V = 2π [(2/3)(2√2) - (2/3)(0)]
V = 2π (4/3)√2
Therefore, the volume of the solid generated by rotating the area bounded by the curve y = x^2, the x-axis, and the lines x = 0 and y = 2 about the y-axis is (8π/3)√2.