1) Volume created when the area bounded by the x-axis, the y-axis, and the line y = -3x+5 is rotated about:
a) the y-axis
b) the line y = 6
2) Volume created when the area bounded by x = 3y-y^2 and the y-axis is rotated about the x-axis
#1
(a) discs ∫[0,5] π((5-y)/3)^2 dy
(a) shells ∫[0,5/3] 2πx(5-3x) dx
(b) shells ∫[0,5] 2π(6-y)(5-y)/3 dy
#2
discs ∫[0,9/4] π(((3+√(9-4x))/2)^2-(((3-√(9-4x)))/2)^2) dx
shells ∫[0,3] 2πy(3y-y^2) dy
#1
(a) 14.544 units^3
(b) 113.446 units^3
#2: 6.75 units^3
is this correct?
I find it easier to leave the π in the answer. That way it's exact, instead of some decimal approximation. In any case, I got
(a) 20π/9
(b) 125π/27
#2 27π/2
To find the volume created when the given areas are rotated, we can use the method of cylindrical shells. Here's how to do it:
1) Volume created when the area bounded by the x-axis, the y-axis, and the line y = -3x+5 is rotated about:
a) the y-axis:
First, let's find the limits of integration. To do this, we need to find the x-values where the line y = -3x+5 intersects the x-axis. Setting y = 0, we have:
0 = -3x + 5
Solving for x, we get:
3x = 5
x = 5/3
Since the line intersects the x-axis at x = 5/3, our limits of integration for x will be from 0 to 5/3.
Next, let's consider a thin vertical strip of width dx at a distance x from the y-axis. The height of this strip can be found by subtracting the y-coordinate of the line y = -3x + 5 from 0 (the y-axis). So the height of the strip is -(-3x + 5) = 3x - 5.
Now, for a given x, the circumference of the cylindrical shell is given by 2πx (since we are rotating about the y-axis). The length of the shell (the width of the strip) is dx. Therefore, the volume of the shell is 2πx * (3x - 5) * dx.
To find the total volume, we integrate the volume of the shell from x = 0 to x = 5/3:
V = ∫[0 to 5/3] 2πx * (3x - 5) dx
Solving this integral will give us the volume.
b) the line y = 6:
To find the limits of integration, we need to find the x-values where the line y = -3x + 5 intersects the line y = 6. Setting these two equations equal to each other, we get:
-3x + 5 = 6
Solving for x, we have:
-3x = 1
x = -1/3
Since the line intersects y = 6 at x = -1/3, our limits of integration for x will be from -1/3 to 5/3.
Following the same steps as in part a), we find the height of the strip as 6 - (-3x + 5) = 11 + 3x.
The volume of the shell is then 2πx * (11 + 3x) * dx, and we integrate this from x = -1/3 to x = 5/3 to find the total volume.
2) Volume created when the area bounded by x = 3y - y^2 and the y-axis is rotated about the x-axis:
First, let's find the limits of integration. To do this, we need to find the y-values where the parabola x = 3y - y^2 intersects the y-axis. Setting x = 0, we have:
0 = 3y - y^2
Simplifying, we get:
y^2 - 3y = 0
y(y-3) = 0
So the parabola intersects the y-axis at y = 0 and y = 3. Our limits of integration for y will be from 0 to 3.
Now, consider a thin horizontal strip of height dy at a distance y from the x-axis. The width of this strip can be found by subtracting the x-coordinate of the parabola x = 3y - y^2 from 0 (the y-axis). So the width of the strip is -(3y - y^2) = y^2 - 3y.
The circumference of the cylindrical shell is given by 2πy (since we are rotating about the x-axis), and the length of the shell (the height of the strip) is dy. Therefore, the volume of the shell is 2πy * (y^2 - 3y) * dy.
To find the total volume, we integrate the volume of the shell from y = 0 to y = 3:
V = ∫[0 to 3] 2πy * (y^2 - 3y) dy
Evaluating this integral will give us the volume.