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

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Answer 1 · 2016-04-16T09:34:15

#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.