# Math calculus by rotation

Calculate the volume obtained by rotating around y= -3 the area between the curves
x= y^2 -4
x+y+2=0

I did the Volume as an integral between (-4, 0) of
(x^2 +x - 12 - 6(x+4) ^1/2) dx

The method I used is seeing the solid as a washer(?
Can someone help me to verify if the answer I have is correct

PD= I have another slight question, why in some formulas for the volume of a rotating solid appears 2 Pi?

Thank you!

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1. when you use discs, the volume is base*height = πr^2h
when you use shells, the volume is circumference*height*thickness = 2πrh dy
So, for this problem, using shells of thickness dy, we have
v = ∫[-3,2] 2πrh dy = ∫[-3,2] 2π(y+3)((2-y)-(y^2-4)) dy = 625π/6

using discs of thickness dx is a bit trickier, since at x=0, the upper boundary changes from the parabola to the line.

v = ∫π(R^2-r^2) dx
v = ∫[-4,0] π((3+√(x+4))^2-((3-√(x+4))^2 dx + ∫[0,5] π((3+(2-x))^2-((3-√(x+4))^2) dx
= 64π + 241π/6 = 625π/6

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