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What is the volume of the solid produced by revolving #f(x)=sqrt(1-x), x in [0,1] #around the x-axis?

Answer 1

#pi/2 cubic units.

Here, #x<=1#. i have changed the interval of integration to [0, 1] from the inadmissible [0, 4]. Of course, negative lower limits are admissible.

The volume =#piinty^2 dx=piint(1-x) dx = pi[x-x^2/2]#, between the limits [0, 1]. answer = #pi/2#, cubic units.

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Answer 2

Ezekiel Alexander

The volume of the solid produced by revolving ( f(x) = \sqrt{1-x} ) over the interval ([0,1]) around the x-axis can be found using the disk method or the washer method in calculus. The volume can be calculated by integrating the area of the cross-sections perpendicular to the x-axis.

Using the disk method, the volume is given by the integral:

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

where ( f(x) = \sqrt{1-x} ), ( a = 0 ), and ( b = 1 ).

Substitute ( f(x) ) and the limits of integration into the formula and compute the 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.