What is the volume of the solid produced by revolving #f(x)=1/x, x in [1,4] #around the x-axis?
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To find the volume of the solid produced by revolving the function ( f(x) = \frac{1}{x} ) around the x-axis over the interval [1, 4], we can use the method of cylindrical shells. The formula for the volume of a solid generated by revolving a function around the x-axis using cylindrical shells is:
[ V = 2\pi \int_{a}^{b} xf(x) , dx ]
Where:
- ( a ) and ( b ) represent the limits of integration.
- ( f(x) ) is the given function.
- ( x ) represents the variable of integration.
Substituting the given function ( f(x) = \frac{1}{x} ) and the limits of integration ( a = 1 ) and ( b = 4 ) into the formula, we get:
[ V = 2\pi \int_{1}^{4} x\left(\frac{1}{x}\right) , dx ]
[ V = 2\pi \int_{1}^{4} 1 , dx ]
[ V = 2\pi \left[\frac{x^2}{2}\right]_{1}^{4} ]
[ V = 2\pi \left(\frac{4^2}{2} - \frac{1^2}{2}\right) ]
[ V = 2\pi \left(\frac{16}{2} - \frac{1}{2}\right) ]
[ V = 2\pi \left(\frac{15}{2}\right) ]
[ V = \pi \cdot 15 ]
[ V = 15\pi ]
Therefore, the volume of the solid produced by revolving the function ( f(x) = \frac{1}{x} ), ( x ) in the interval [1,4], around the x-axis is ( 15\pi ).
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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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