R is the area bounded by f(x) = (x^3)/4)- (x^2)/3 - x/2) +3cos x.
Find the volume of the solid generated when R is rotated about the horizontal line y = -2.

I got Volume = pi times the integral from -1.373 to 0 of (f(x)+2)^2) dx.
Or is it supposed to be (f(x)+2)^2 - 2^2)?

Question

R is the area bounded by f(x) = (x^3)/4)- (x^2)/3 - x/2) +3cos x.
Find the volume of the solid generated when R is rotated about the horizontal line y = -2.

I got  Volume = pi times the integral from -1.373 to 0 of (f(x)+2)^2) dx.
Or is it supposed to be (f(x)+2)^2 - 2^2)?

anonymous · Answer

If you find the intersecting points between the Horizontal line y = -2
and the graph f(x) you get only one intersecting point at -2.54 =x. The rest of the function f(x) is unbounded I need more information the region betwen the line y=-2 and the graph f(x) = (x^3)/4)- (x^2)/3 - x/2) +3cos x is not! fully enclosed???

anonymous · Answer

The f(x) is bounded from the x-axis to the y-axis. That is used as the base to rotate it around y = -2.

anonymous · Answer

So  the the Y-axis is indeed the additional given axis bounding the region between f(x) and the Line y = -2. Ok, i did not see that in the original question ok.

anonymous · Answer

Then the additional point is where the value of x when f(x) crosses the y-axis ( 0 ). Hence [ -2.54, 0 ] are the lower and upper limits of integration. You can just use the disc method to solve. ......................pi*Integration[ (f(x)+ 2)^2, -2.54,0 dx ]


Done.