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Graphs of x=f(y)=y(3−2y) and x=0 enclose a region in the first quadrant. Rotating that region about the x-axis generates a solid whose volume is...

User Bud
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Answer:

To find the volume of the solid generated by rotating the region enclosed by the graphs of x = f(y) = y(3 - 2y) and x = 0 about the x-axis, we can use the method of cylindrical shells.

The first step is to set up the integral that represents the volume of the solid. Since we are rotating the region about the x-axis, the radius of each cylindrical shell will be y, and the height will be the difference between the x-values of the curves at each y-value.

To find the limits of integration, we need to determine the y-values at which the two curves intersect. Setting y(3 - 2y) = 0, we find that y = 0 and y = 1.5 are the solutions.

Now we can set up the integral:

V = ∫[a,b] 2πy * [f(y) - 0] dy

where [a, b] represents the interval of integration, which in this case is [0, 1.5].

Simplifying the integral:

V = ∫[0,1.5] 2πy * y(3 - 2y) dy

V = 2π ∫[0,1.5] y^2 (3 - 2y) dy

Evaluating the integral:

V = 2π [∫[0,1.5] (3y^2 - 2y^3) dy]

V = 2π [y^3 - (2/4)y^4] evaluated from 0 to 1.5

V = 2π [(1.5)^3 - (2/4)(1.5)^4] - 2π [0^3 - (2/4)(0)^4]

V = 2π [(1.5)^3 - (2/4)(1.5)^4]

V = 2π [3.375 - (2/4)(3.375)]

V = 2π [3.375 - 1.6875]

V = 2π [1.6875]

V = 3.375π

Therefore, the volume of the solid generated by rotating the region enclosed by the graphs of x = f(y) = y(3 - 2y) and x = 0 about the x-axis is 3.375π cubic units.

User Dsbisht
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