Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the graphs of the given equations about the y-axis.y = 5x2,    y = 30x − 10x2

Sure, here is the step-by-step solution:

Step 1: Identify the region bounded by the graphs The region bounded by the graphs y = 5x^2 and y = 30x - 10x^2 is a region in the first quadrant of the xy-plane.

Step 2: Set up the integral for the volume The volume V of the solid obtained by rotating the region about the y-axis is given by the integral:

V = ∫[a,b] 2πxh dx

where [a,b] is the interval of x-values that the region spans, x is the radius of the cylindrical shell at a given x-value, and h is the height of the cylindrical shell at that x-value.

Step 3: Determine the radius and height of the cylindrical shells The radius of the cylindrical shell at a given x-value is just x. The height of the cylindrical shell at that x-value is the difference between the y-values of the two functions at that x-value, which is (30x - 10x^2) - 5x^2 = 30x - 15x^2.

Step 4: Evaluate the integral Substitute the expressions for the radius and height into the integral and evaluate it to find the volume. The limits of integration are the x-values where the two functions intersect, which can be found by setting the two functions equal to each other and solving for x:

5x^2 = 30x - 10x^2 15x^2 = 30x x = 30/15 = 2

So the limits of integration are 0 and 2. The volume is then:

V = ∫[0,2] 2πx(30x - 15x^2) dx = 2π ∫[0,2] (30x^2 - 15x^3) dx = 2π [10x^3 - 15/4 x^4] evaluated from 0 to 2 = 2π [80 - 60] = 40π cubic units.