- #1
stephencormac
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Can anyone confirm if I have done the following work correctly
Find the volume of a solid of revolution obtained by rotating about the y-axis the region bounded by y = the fifth root of x and 2x^2 - 3x + 2.
By drawing the graph, I figured out that I need to use the method of cylindrcal shells given by v = integral from 0.619 to 1 of A(x) dx.
Where A(x) = 2pi(radius)(height)
The intersection points of the equations are approximately x = 1 and x = 0.619
Radius is equal to x.
Height is equal to the difference in the two equations
i.e. (x^(1/5) - 2x^2 + 3x - 2)
Thus we have 2pi(radius)(height)
=2pi*(x)*(x^(1/5) - 2x^2 + 3x - 2)
= 2pi*(x^(6/5) - 2x^3 + 3x^2 - 2x)
Now I will integrate this between 0.619 and 1
= (5/11)x^(11/5) - (1/2)x^4 + x^3 - x^2 ¦ 0.619 to 1
which gives me 2pi*(-0.05 + 0.1345)
=0.169pi approximately.
Can anyone confirm that I have done this correctly
Find the volume of a solid of revolution obtained by rotating about the y-axis the region bounded by y = the fifth root of x and 2x^2 - 3x + 2.
By drawing the graph, I figured out that I need to use the method of cylindrcal shells given by v = integral from 0.619 to 1 of A(x) dx.
Where A(x) = 2pi(radius)(height)
The intersection points of the equations are approximately x = 1 and x = 0.619
Radius is equal to x.
Height is equal to the difference in the two equations
i.e. (x^(1/5) - 2x^2 + 3x - 2)
Thus we have 2pi(radius)(height)
=2pi*(x)*(x^(1/5) - 2x^2 + 3x - 2)
= 2pi*(x^(6/5) - 2x^3 + 3x^2 - 2x)
Now I will integrate this between 0.619 and 1
= (5/11)x^(11/5) - (1/2)x^4 + x^3 - x^2 ¦ 0.619 to 1
which gives me 2pi*(-0.05 + 0.1345)
=0.169pi approximately.
Can anyone confirm that I have done this correctly