# green theorem using green theorem for solving this questionclose integral of (y^3*dx+x^3*dy) while x^2+y^2=4 the final answer is 12pi,but i dont know why my answer is different

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You need to remember the equation given by the Green's theorem, that relates a positively oriented closed curve C and the partail derivatives of the functions P and Q, such that:

`int_C (Pdx + Qdy) = int int_D ((delQ)/(delx) + (delP/(dely))) dA`

Notice that the problem provides the following equation that described the curve,`x^2 + y^2 = 4` , hence, the curve is a circle, centered at origin and of radius 2.

Using Green's theorem yields:

`int_C (y^3dx + x^3dy) = int int_D (((del(x^3)))/(delx) + ((del(y^3))/(dely))) dA`

`int_C (y^3dx + x^3dy) = int int_D (3x^2 + 3y^2) dA`

Since D is a disk of radius 2, centered at origin, yields:

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) int_0^2 (x^2 + y^2) dr d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) int_0^2 (r^2cos^2 theta + r^2 sin^2 theta) dr d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) int_0^2 r^2(cos^2 theta + sin^2 theta) dr d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) int_0^2 r^2*r dr d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) int_0^2 r^3 dr d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) r^4/4|_0^2 d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) (16/4 - 0/4) d theta`

`int_C (y^3dx + x^3dy) = 3int_0^(2pi) 4 d theta`

`int_C (y^3dx + x^3dy) = 12int_0^(2pi) d theta`

`int_C (y^3dx + x^3dy) = 12 theta|_0^(2pi)`

`int_C (y^3dx + x^3dy) = 12 (2pi - 0)`

`int_C (y^3dx + x^3dy) = 24pi`

**Hence, evaluating the closed integral using Green's theorem, yields `int_C (y^3dx + x^3dy) = 24pi.` **