Asked by Claire
Let R be the region between the graph of y=x3 and the x-axis, and between x=0 and x=1.
Compute the volume of the region obtained by revolving R around the y-axis
Compute the volume of the region obtained by revolving R around the y-axis
Answers
Answered by
Steve
with discs (washers),
v = ∫[0,1] π (R^2-r^2) dy
where r=x and R=1
v = π∫[0,1] (1 - y^(2/3)) dy
= π (y - 3/5 y^(5/3)) [0,1]
= 2/5 π
with shells,
v = ∫[0,1] 2π rh dx
where r=x and h=y
v = 2π∫[0,1] x*x^3 dx
= 2π (1/5 x^5) [0,1]
= 2/5 π
v = ∫[0,1] π (R^2-r^2) dy
where r=x and R=1
v = π∫[0,1] (1 - y^(2/3)) dy
= π (y - 3/5 y^(5/3)) [0,1]
= 2/5 π
with shells,
v = ∫[0,1] 2π rh dx
where r=x and h=y
v = 2π∫[0,1] x*x^3 dx
= 2π (1/5 x^5) [0,1]
= 2/5 π
Answered by
Claire
First off, thank you answering my question.
I just have one question. So does it matter which method we use?
I am just trying to find a better way of understanding this problem from a visual perspective.
I just have one question. So does it matter which method we use?
I am just trying to find a better way of understanding this problem from a visual perspective.
Answered by
Steve
generally it doesn't matter. It depends on whether it is easier to express y as a function of x, or vice-versa. For example if y=e^x + 4(x^2-1) then it's pretty tough to come up with y<sup>-1</sup>
In this case, it was easy.
In this case, it was easy.
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