Intersection of Plane and Paraboloid and Triple Integral

Let W be the region below the paraboloid x^2+y^2=z-2 that lies above the part of the plane x+y+z=1 in the first octant.

I need to calculate the triple integral to find the volume of the region.
I set the plane and paraboloid equal and keep getting the intersection as a circle with negative radius when i complete the squares (x+1/2)^2+ (y+1/2)^2= -1/2. Am i making a mistake? Of course the integral for the z part is from the plane to the paraboloid, but i am unsure how to find the limits for the x and y parts of the integral because of the negative radius.

your work tells you that they DO NOT intersect { in the 1st quadrant }...

thus int over x in [ 0 , 1 ] , y in [ 0 , 1 - x ] of { (x² + y² + 2 ) - ( 1 - x - y ) } dydx

in the x - y plane you have a triangular region for the domain of integration...then top - bottom

Just a few points of clarification.
If a circle has negative radius, as in this case, is the circle still centered at what it would be if the radius were positive. So in this case, would the center be (-1/2, 1/2) still?

Report Abuse

Also if the two z functions do not intersect, as in this case, do we always use the bottom function and set it equal to 0 to find the domain on the xy plane?

Thanks for the help

Report Abuse