finding volume through integration?

Question

the graph of f(x,y) = 4xy over the bounded region A in the first quadrant enclosed by y = sq.rt.(9-x^2) and the x, y-axes is the surface. find the volume of the solid under this graph over the region A.

Answer 1

The area A is the first quadrant of the circle x^2 + y^2 = 9.

Set it up as a double integral, and integrate on y before x:

V = int(0 to 3) int[0 to sqrt(9 - x^2)] f(x,y) dy dx
V = int(0 to 3) int[0 to sqrt(9 - x^2)] (4xy) dy dx
V = int(0 to 3) (2xy^2) [0 to sqrt(9 - x^2)] dx
V = int(0 to 3) (2x)(9 - x^2) dx
V = int(0 to 3) (18x - 2x^3) dx
V = (9x^2 - 1/2 x^4) (0 to 3)
V = 81 - 81 / 2 = 81 / 2 = 40.5 (answer)

Answer 2

think that the based hollow is vertical .... make x-axis on center (horizontal) u ought to make double integration ..........2*INT( int (11^2-X^2-Y^2) )...... get this equation via understanding equation of ball x^2+y^2+z^2=11^2 and placed the two bec i'm utilising from ball to the x-axis (abovex-axis section) so the is a factor under it so i multipied via 2 now u ought to alter X=Rcos(ceta) ...Y=Rsin(ceta) the equation would be 2*INT(int(11^2-r^2)) ... bec sin^2+cos^2 = a million the 1st int would be from R=11 to R=9(with understand to r) ...... and the 2nd from ceta=360 to ceta=0( with understand to ceta ) ..... wish U understand ME