shell method y=5x^2, Y=30x-10x^2 about y axis

Question

i do not no what to do

Answer 1

We have two intersecting parabolas y=5x² and y=10x-30x², so the first task is to find out where they intersect. When 5x²=30x-10x², they intersect, so we need to solve for x.

15x²-30x=0=15x(x-2). So x=0 and 2. The intersection points are (0,0) and (2,20). Between these points we have an enclosed area, and it's this area which is to be rotated around the y-axis, which means that the radius of the cylindrical shells is going to be the x value. Consider the volume of a cylinder: πr²h where r is the radius and h the height of the cylinder. In this problem we've already identified the radius as the x value. Since the height is perpendicular to the radius, the height is going to depend on the y value. But we have two equations for y.

The difference in y values for some particular value of x is |5x²-(30x-10x²)|=|15x²-30x|. For example, when x=½, the height of the cylinder is |15/4-15|=|-45/4|=45/4. Note that we need the height to be positive so we need to negate 15x²-30x to give us a positive value: 30x-15x². On a graph you would be able to see that in the x interval [0,2] the graph of y=30x-10x² lies above y=5x², so that would tell you the positive height was 30x-10x²-5x²=30x-15x². It helps to sketch a graph.

If we consider two cylinders of different radii (R and r) but the same height h, then the difference in volume is the volume of the cylindrical shell=πh(R²-r²).

If R=r+dr, then this becomes πh((r+dr)²-r²)=πh(2rdr+(dr)²) and if dr is very small it can be ignored, so the difference becomes 2πrhdr. Since radius is the x value in this problem this is 2πxhdx. If we consider only one of the curves we can work out the volume of that curve rotated around the y-axis as an integral. Let's use y=5x² as an example. We are measuring the volume between x=0 and x=2, so we have:

∫2πxydx=∫2πx(5x²)dx=∫10πx³dx for 0≤x≤2.

If we now find the volume of the other curve we get: ∫2πx(30x-10x²)dx=∫(60πx²-20πx³)dx for 0≤x≤2.

The difference between these two volumes is the volume we need, so we can combine the two integrals into one:

∫(60πx²-20πx³-10πx³)dx=∫(60πx²-30πx³)dx=∫2πx(30x-15x²)dx. Note that 30x-15x² is the difference in the y values of the two functions.

The limits for the definite integration are 0 and 2. Volume of rotation=[20πx³-15πx⁴/2]₀²=160π-120π=40π cubic units.