Confusing integration question

not sure how to put the question on here but yeah its like V = int int R (x^2 + xy^3)dA and R is 0 less than or equal to x less than or equal to 1. And 1 less than or equal to y less than or equal to 2.

double integration by the way

http://oi49.tinypic.com/i5uo14.jpg

All my working out

That's about it!

If I understand the equation, it is

V = dbl int(x^2 + xy^3) dxdy, x = [0, 1], y = [1, 2]

If this is the case, then break the integral into two parts:

V = dbl int(x^2) dxdy + dbl int(xy^3) dxdy, x = [0, 1], y = [1, 2]

The indefinite integral of x^2 is x^3 / 3, so

dbl int(x^2) dxdy is (x^3/3) on the interval [0, 1], or

1^3/3 - 0^3/3 = 1/3 - 0 = 1/3

The first integral becomes

int(1/3) dy = y/3, on the interval [1, 2]

Calculating, int(1/3) dy = 2/3 - 1/3 = 1/3

Therefore, the first part of the integral is 1/3.

Moving onto the second part, int(xy^3) dxdy, we integrate with respect to x, so it is the same as

int(cx) dx = x^2 y^3 / 2, on the interval x = [0, 1]

Evaluating,

int(xy^3) dx = 1^2 y^3 / 2 - 0^2 y^3 / 2 = y^3 / 2 - 0 = y^3 / 2

Now,

V = 1/3 + int(y^3 / 2) dy, y = [1, 2]

Integrating with respect to y gives

V = 1/3 + y^4 / 8, y = [1, 2]

V = 1/3 + (2^4 / 8 - 1^4 / 8)

V = 1/3 + 16/8 - 1/8

V = 1/3 + 15/8

V = 8/24 + 45/24

V = 53 / 24

∫∫ (x^2 + xy^3) dx dy 0 ≤ x ≤ 1 and 1≤ y ≤ 2

= ∫ [1/3(x^3 + 1/2x^2y^3) from ( 0 to 1 ) dy

= ∫ [(1/3) + 1/2y^3) dy

= (1/3) y + (1/8) y^4 (from 1 to 2 )

= (1/3)(2 - 1) + (1/8)(16 - 1)

= (1/3) + (15/8)

= 53/24