How do you find the volume of a solid that is enclosed by #y=-x^2+1# and #y=0# revolved about the x-axis?

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Answer 1 · 2017-11-10T16:47:49

#~~3.3510322# cubic units.

#-x^2+1=0=>x=1 or x=-1#

#-x^2+1#

is the derivative of some area function. We now need to integrate the square of this, since we are looking for the volume.

#V=pi*int_-1^(1)(-x^2+1)^2 dx#

It will be easier to integrate if we multiply out the function first.

#(-x^2+1)^2=x^4-2x^2+1#

#V=pi*int_-1^(1)(x^4-2x^2+1) dx=x^5/5-(2x^3)/3+x#

#V=pi*[x^5/5-(2x^3)/3+x]_-1^1#

plugging in values for upper and lower bounds.

#[(1)^5/5-(2(1)^3)/3+(1)]-[(-1)^5/5-(2(-1)^3)/3+(-1)]#

#V=pi*[1/5-2/3+1]-[-1/5+2/3-1]=[16/15]#

#V=pi*[16/15]=16/15*picolor(blue)(~~3.3510322)# cubic units.

Plot of revolution:

enter image source here