The region between the graphs of x=y^2 and x=2y is rotated around the line y=2.
The volume of the resulting solid is ? Please show a step by step solution. Thank you
The volume of the resulting solid is ? Please show a step by step solution. Thank you
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I'll use the washer method and integrate with respect to x. First if you draw a graph the two functions, you'll see that they intercept at x=0 and x=4. These will be the limits of integration.
The outer radius of the washer will be 2-1/2x and the inner radius will be 2-sqrt(x).
The area function A(x) = π [(2-1/2x)^2 - (2-sqrt(x))^2] (i.e., because the area of a circle is πr^2)
Expanding this yields: π [x^2/4 - 3x +4*sqrt(x)] (you could have integrate the above but I find it easier to expand first to get simple terms)
So V= ∫ A(x)dx between x=0 and x=4 or
=π [x^3/12 - 3x^2/2 + 8x^(3/2)/3] between 4 and zero or
=π [64/12 - 48/2 + 64/3] = 8π/3
The outer radius of the washer will be 2-1/2x and the inner radius will be 2-sqrt(x).
The area function A(x) = π [(2-1/2x)^2 - (2-sqrt(x))^2] (i.e., because the area of a circle is πr^2)
Expanding this yields: π [x^2/4 - 3x +4*sqrt(x)] (you could have integrate the above but I find it easier to expand first to get simple terms)
So V= ∫ A(x)dx between x=0 and x=4 or
=π [x^3/12 - 3x^2/2 + 8x^(3/2)/3] between 4 and zero or
=π [64/12 - 48/2 + 64/3] = 8π/3