Find the volume bounded above by parabolic z = 6 - x^2 - y^2 and below by z= 2x^2 + y^2

Answer 1

The paraboloids intersect along an ellipse

3x^2+2y^2 = 6

So, the volume is (using symmetry)

v = 4∫[0,√2] ∫[0,√((6-3x^2)/2)] (6-x^2-y^2)-(2x^2+y^2) dy dx = 3π√6

Answer 2

Evaluate double integral ln((x-y)/(x+y)) dy dx where the double integral region is the triangle with vertices (1,0),(4,3), (4,1). Hint: use a transformation with the Jacobian.