# How do you find the volume of the region bounded by #y=sqrt x#, y=0, x=0, and x=4 is revolved about the x-axis?

The volume of the small strip of width dx is given by

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To find the volume of the region bounded by ( y = \sqrt{x} ), ( y = 0 ), ( x = 0 ), and ( x = 4 ) when revolved about the x-axis, you can use the method of cylindrical shells.

The formula for the volume of a solid generated by revolving a region bounded by ( y = f(x) ), ( y = 0 ), ( x = a ), and ( x = b ) about the x-axis is:

[ V = 2\pi \int_{a}^{b} x \cdot f(x) , dx ]

Using this formula, we can calculate the volume for the given region:

[ V = 2\pi \int_{0}^{4} x \cdot \sqrt{x} , dx ]

[ V = 2\pi \int_{0}^{4} x^\frac{3}{2} , dx ]

[ V = 2\pi \left[ \frac{2}{5}x^\frac{5}{2} \right]_{0}^{4} ]

[ V = 2\pi \left( \frac{2}{5} \cdot 4^\frac{5}{2} - \frac{2}{5} \cdot 0^\frac{5}{2} \right) ]

[ V = 2\pi \left( \frac{2}{5} \cdot 8 - 0 \right) ]

[ V = 2\pi \left( \frac{16}{5} \right) ]

[ V = \frac{32}{5} \pi ]

So, the volume of the region bounded by ( y = \sqrt{x} ), ( y = 0 ), ( x = 0 ), and ( x = 4 ) when revolved about the x-axis is ( \frac{32}{5} \pi ) cubic units.

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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.

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