How do you find the volume of the solid generated by revolving the region bounded by the graphs #y=e^(x/2), y=0, x=0, x=4#, about the x axis?
Please see below.
Here is a graph of the region in blue. A slice has been taken perpendicular to the axis of rotation. The rotation as shown by the arrow/arc.
The representative slice is a disc of
thickness dx
and radius
The volume of the representative slice (disc) is
The values of
Evaluate the integral to get
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To find the volume of the solid generated by revolving the region bounded by the graphs ( y = e^{x/2} ), ( y = 0 ), ( x = 0 ), and ( x = 4 ) about the x-axis, you can use the method of cylindrical shells.
The formula for the volume of the solid generated by revolving the region bounded by the graphs of two functions ( y = f(x) ) and ( y = g(x) ) (where ( f(x) ) is the upper function and ( g(x) ) is the lower function) about the x-axis from ( x = a ) to ( x = b ) is given by:
[ V = \int_{a}^{b} 2\pi x (f(x) - g(x)) , dx ]
In this case, ( f(x) = e^{x/2} ), ( g(x) = 0 ), ( a = 0 ), and ( b = 4 ).
So, the volume ( V ) is:
[ V = \int_{0}^{4} 2\pi x (e^{x/2} - 0) , dx ]
You can then integrate this expression to find the volume of the solid.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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