Find the volume of the solid obtained by rotating the region enclosed by the graphs of y=15-x, y=3x+11 and x=0 about the y-axis.
To find the volume of the solid obtained by rotating the region enclosed by the given graphs, we can use the method of cylindrical shells. This involves summing up the volumes of infinitely thin cylindrical shells that make up the solid.
First, let's visualize the region enclosed by the graphs of the given equations:
The equations are:
y = 15 - x
y = 3x + 11
x = 0
To find the boundaries of the region, we need to determine the points where the curves intersect.
Setting the two equations y = 15 - x and y = 3x + 11 equal to each other:
15 - x = 3x + 11
Simplifying the equation:
4x + 11 = 15
Solving for x:
4x = 4
x = 1
Substituting the value of x back into either equation, we find that y = 14.
So the region is enclosed between x = 0 and x = 1, and bounded by y = 15 - x and y = 3x + 11.
To set up the integral for finding the volume, imagine a horizontal slice at a distance x from the y-axis. The width of this slice is dx, and its height is the difference between the two curves at that particular x-value.
The height of the cylindrical shell is the difference between the two curves:
h = (15 - x) - (3x + 11)
h = 4 - 4x
The radius of the cylindrical shell is the distance from the y-axis to that particular x-value, which is x.
The volume of the cylindrical shell is given by the formula:
dV = 2πrh dx
Substituting the values into the formula, we get:
dV = 2π(x)(4 - 4x) dx
Now, we can integrate the above expression over the interval [0, 1] to obtain the total volume:
V = ∫(0 to 1) 2π(x)(4 - 4x) dx
Integrating this expression:
V = 2π ∫(0 to 1) (4x - 4x^2) dx
V = 2π [2x^2 - (4/3)x^3] from 0 to 1
Substituting the limits of integration:
V = 2π [(2(1)^2 - (4/3)(1)^3) - (2(0)^2 - (4/3)(0)^3)]
Simplifying the expression:
V = 2π [(2 - 4/3) - (0)]
V = 2π (6/3 - 4/3)
V = 2π (2/3)
Finally, simplifying:
V = 4π/3
Therefore, the volume of the solid obtained by rotating the region enclosed by the given graphs about the y-axis is 4π/3 cubic units.
Find out where the curves y = 15-x and y = 3x + 11 intersect
15 -x = 3x + 11.
4x = 4
x = 1. At that point, y = 14.
The enclosed volume goes from the y axis to x = 1. You are rotating a triangle about the y axis. The limits of integration are from y = 11 to y = 15. Those are the maximum and minimum y values of the enclosed volume.
Do an integration of
2 pi x^2 dy from y=11 to y=14, using the y = 3x + 11 function, and add the integral of 2 pi x^2 dy from y=14 to y = 15, using the y = 15 -x function. You will have to convert x to y in those integrations, using the appropriate formula.
It's a lot easier to visualize if you draw a figure, but I can't do that here.