How do you find the volume of a rotated region bounded by #y=sqrt(x)#, #y=3#, the y-axis about the y-axis?

Answer 1

Christopher Allers

Try this:

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Answer 2

Christopher Allers

Circular cross sections of the bounded region have an area
#pi x^2#
or, since #x = y^2#
#A(y) = pi y^4#

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Answer 3

Charles Anctil

To find the volume of the rotated region bounded by ( y = \sqrt{x} ), ( y = 3 ), and the y-axis about the y-axis, you can use the method of cylindrical shells. The formula for finding the volume using cylindrical shells is ( V = 2\pi \int_{a}^{b} xf(x) , dx ), where ( f(x) ) represents the height of the shell at a given x-value, and ( a ) and ( b ) represent the limits of integration.

In this case, the height of the shell is the difference between the outer radius (which is the distance from the y-axis to the function ( y = 3 )) and the inner radius (which is the distance from the y-axis to ( y = \sqrt{x} )). So, ( f(x) = 3 - \sqrt{x} ).

To find the limits of integration, you need to determine where the curves intersect. Set ( \sqrt{x} = 3 ) and solve for ( x ) to find the x-coordinate of the intersection point. Then, integrate from ( x = 0 ) to that x-coordinate.

The volume can be calculated as follows:

[ V = 2\pi \int_{0}^{9} x(3 - \sqrt{x}) , dx ]

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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.