find the volume under the plane z= x+y that is over the rectangular region R:0<=

Question

find the volume under the plane z= x+y that is over the rectangular region R:0<=x<=2 0<=y<=1

Explanation / Answer

1) As for z, these form the integrand; note that 3x > x^2 for all x in (0, 3). So, the integral is given by ?(y = 0 to 4) ?(x = 0 to 4) (3x - x^2) dx dy = ?(y = 0 to 4) (3x^2/2 - x^3/3) {for x = 0 to 4} dy = ?(y = 0 to 4) (8/3) dy = 4 * 8/3 = 32/3. ------------------- 2) Note that the vertices of the triangle are (0, 0), (4, 0), and (0, 2). (Graphing this may be useful; the last two vertices are the intercepts of x + 2y = 4.) The integral can be set up either way; I'll choose the way without fractions: x = 0 to x = 4 - 2y with y in [0, 2]. Hence, the integral equals ?(y = 0 to 2) ?(x = 0 to 4 - 2y) xy dx dy = ?(y = 0 to 2) (1/2)x^2 y {for x = 0 to 4 - 2y} dy = ?(y = 0 to 2) (1/2)(4 - 2y)^2 y dy = ?(y = 0 to 2) (2y^3 - 8y^2 + 8y) dy = (y^4/2 - 8y^3/3 + 4y^2) {for y = 0 to 2} = 8/3.

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