How do you find the volume of the solid generated when the regions bounded by the graphs of the given equations #y = sqrt (3 - x^2)#, x=0, x=1 and the x-axis are rotated about the x-axis?

Answer 1

#(8pi)/3# cubic units.

The given equation represents the circle #x^2+y^2=3

Volume =#piint y^2 dx#, from x=0 t0 x=1

#pi[3x-x^3/3], beteween x = 0 and x = 1

#=pi(3-1/3)#
#=pi(8/3)#.-

The solid generated is a slice of a hemisphere of radius sqrt 3 units,

between the base abd the parallel plane,at a distance 1 unit.i

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Answer 2

To find the volume of the solid generated when the regions bounded by the graphs of the equations (y = \sqrt{3 - x^2}), (x = 0), (x = 1), and the x-axis are rotated about the x-axis, you can use the method of cylindrical shells. The formula for the volume of the solid generated by rotating the region between the curve (y = f(x)), the lines (x = a) and (x = b), and the x-axis about the x-axis is given by:

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

In this case, (f(x) = \sqrt{3 - x^2}), (a = 0), and (b = 1). Therefore, the volume can be calculated using the given formula.

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Answer from HIX Tutor

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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