How do you find the volume of a solid that is enclosed by #y=sqrt(4+x)#, x=0 and y=0 revolved about the x axis?

Answer 1

The volume #=8pi#

The volume of a disc of thickness dx is #dV=piy^2dx# as #y=sqrt(4+x)# #=># #dV=pi(4+x)dx# The limits of integration are #x=0# and #x=-4# As #4+x>=0# so the volume #V=int_-4^0pi(4+x)dx=pi(4x+x^2/2)_-4^0# #=-pi(-16+8)=8pi# graph{sqrt(x+4) [-10, 10, -5, 5]}
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Answer 2

To find the volume of the solid formed by revolving the region enclosed by the curve ( y = \sqrt{4 + x} ), the x-axis, and the line ( x = 0 ) about the x-axis, we can use the method of cylindrical shells.

The formula to find the volume using cylindrical shells is:

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

Where ( f(x) ) represents the height of the shell at each value of ( x ), and ( [a, b] ) represents the interval of integration.

In this case, the interval of integration is from ( x = 0 ) to the x-coordinate where ( y = 0 ), which is the x-intercept of the curve. Solving ( y = \sqrt{4 + x} = 0 ) gives ( x = -4 ).

Thus, the integral becomes:

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

Solve this integral to find the volume.

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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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