I need step by step the procedure to reach the solution, please write it clear

1 Exercise

Calculate 11, ( x2 - > ) dx dy, Where D is the region between the graph of the parabolas y = -X , y = x y las rectas x = -1, x = 1. Solution: uilA Resolution. Integration enclosure D is shown in the figure gey 0.5 05 .0.5 It is a type I region. Therefore, we can calculate the requested integral by writing [ ( x2 - y) dxdy = [, dx [ (2 - may = [, xy-2],dx= 2/ xax=3.Draw the region of integration and change the order in the integral [ dx ( 2 f (x,y) dy. Solution Resolution. The region of integration is shown in the figure 0.8 0.6 - 04 - 03. 0.5 6is X Every horizontal line between 0 and 1 enters the domain at an abscissa point * = - Vy and comes out at an abscissa point x = -y , to re-enter an abscissa point x = y and exit at an abscissa point x = Vy. Therefore, the integral is written in the formCalculate the area of the portion of the paraboloid > = " + y that is included between the planes z = 0 y z =1. Soluition: The intersection of the paraboloid with the plane z = 0 is the point (0, 0) and with the plane z = 1 is the circumference r+ y = 1. The region limited by the projection of said circumference on the plane XY is D= {(x,y) ER' : aty's1 ). Z ,2) Y X We can consider the following parameterization: "(T, y) = (x, y, 7 + y'), (x, y) ED. n this way S = r(D), where S is the surface described in the statement. Its fundamental vector product is: N(x, y) = (-2x, -2y, 1), y IN(x, y)| = V4x2 + 4y2 + 1. We will do this integral through a change of variable to polar coordinates T = P COS 4 0 0. y = psen 0