Find the volume of the solid generated by revolving the curves y=x^2+1 & y=x+3.
after finding the intersection, x= 2 & x=-1
can someone show me how to integrate this funct. with simplest steps(i mean do not use substitution formula).
if im x miataken the ans is 117/5 pie unit^3
I assume you want to rotate the region around the x-axis. If not, you will have to adjust things accordingly. Since the x-axis is outside the region, the cross-sections will be washers with a hole in the middle.
So, the volume is
∫[-1,2] π(R^2-r^2) dx
where R = x+3 and r = x^2+1
Now it's just a matter of substituting in for R and r:
π∫[-1,2] ((x+3)^2-(x^2+1)^2) dx
π∫[-1,2] -x^4 - x^2 + 6x + 8 dx
π(-1/5 x^5 - 1/3 x^3 + 3x^2 + 8x) [-1,2]
π[(-32/5-8/3+12+16)-(1/5+1/3+3-8)]
117/5 π
you are correct