Calc: Volume generated by revolving regions bounded by curve.?

LIKE YOU UNDERSTOOD , YES SIR, THAT IS THE INTEGRAL USING CYLINDRICAL COORD. - ^. Z = (x^2 + y^2)/2 , so 2z = (x^2 + y^2) , plug in the other one 2z+z^2=8 Complete squares (z^2+2z+1)-1=8 (z+1)^2 =9 z+1=+-3 z=2 z=-4 So we have points on XY plane such as x^2+y^2 =2z , ie x^2+y^2=4 for z=2 ( Plane z=2) or x^2+y^2= -8 ( No solution , because z=(x^2 + y^2)/2 is a vertical paraboloid open to Z+ upwards , no interception at Z<0) ) So the curve is a circle x^2+y^2=4 , Radius R=2 at a plane z= 2 Use cylindrical coordinates, the Region will be that circle ^.

X=rcosT y=rsinT dA= rdrdT z=z The paraboloid is 2z= r^2 , ie z= r^2/2 The sphere is z^2= 8-r^2 , z= sqrt (8-r^2) dV= dz dA = dz(rdrdT) V= INT_V dz(rdrdT) V=INT_R z rdrdT sqrt (8-r^2)

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