Find the volume of the region under the graph of fxy3xy1 and

Find the volume of the region under the graph of f(x,y)=3x+y+1 and above the region y2?x, 0?x?9.

Solution

1) ?(x = 0 to 1) ?(y = 0 to 3?x) (x^2 + y^2) dy dx
= ?(x = 0 to 1) (x^2 y + y^3/3) {for y = 0 to 3?x} dx
= ?(x = 0 to 1) (x^2 * 3?x + (3?x)^3/3) dx
= ?(x = 0 to 1) (3x^(5/2) + 9x^(3/2)) dx
= [3 * (2/7)x^(7/2) + 9 * (2/5)x^(5/2)] {for x = 0 to 1}
= (6/7 + 18/5) - 0
= 156/35.

?(x = 0 to 3) ?(y = x to 2x) y dy dx
= ?(x = 0 to 3) y^2/2 {for y = x to 2x} dx
= ?(x = 0 to 3) (3/2)x^2 dx
= (1/2)x^3 {for x = 0 to 3}
= 27If f(x,y) = 1, then the area of the rectangle equals the volume under this function f on R.
==> ?(x = 0 to 9) ?(y = 9 to 10) 1 dy dx
= ?(x = 0 to 9) (10 - 9) dx
= (9 - 0)(10 - 9)
= 9.