Answer :
Answer:
The answer is [tex]\pi(\pi-6)[/tex]
Step-by-step explanation:
Recall that green theorem is as follows: Given a field F(x,y) = (P(x,y),Q(x,y)) and a closed curve C that is counterclockwise oriented. If P and Q are continuosly differentiable, then
[tex]\oint_C F\cdot dr = \int_{R} \frac{\partial P }{\partial y}-\frac{\partial P }{\partial x} dA[/tex]
where R is the region enclosed by the curve C.
In this particular case, we have the following field [tex]F(x,y) = (4xy^2,2x^3+y)[/tex]. We are given the description of the region R as [tex]0\leq y \leq \sin(x), 0\leq x \leq \pi[/tex]. So, in this case (calculations are omitted)
[tex]\frac{\partial P}{\partial y} = 8xy, \frac{\partial Q}{\partial x} = 6x[/tex]
Thus,
[tex]\oint_C F\cdot dr =\int_{0}^{\pi}\int_{0}^{\sin(x)}(8xy-6x)dydx[/tex]
So,
[tex] \int_{0}^{\pi}\int_{0}^{\sin(x)}(8xy-6x)dydx=\int_{0}^{\pi}4x\left.y^2\right|_{0}^{\sin(x)}-6x\left.y\right|_{0}^{\sin(x)} = \int_{0}^{\pi} 4x\sin^2(x)-6x\sin(x)dx[/tex]
Since [tex]\sin^2(x) = \frac{1-\cos(2x)}{2}[/tex], then
[tex] \int_{0}^{\pi} 4x\sin^2(x)-6x\sin(x)dx = \int_{0}^{\pi} 2x(1-\cos(2x))-6x\sin(x)dx[/tex]
Consider the integrals
[tex] I_1 = \int_{0}^{\pi} x\cos(2x)dx, I_2 = \int_{0}^{\pi}x\sin(x)dx[/tex]
Then, by using integration by parts (whose calculations are omitted) we get
[tex]\int_{0}^{\pi} x\cos(2x) = \left.\frac{x\sin(2x)}{2}+\frac{\cos(2x)}{4}\right|_{0}^{\pi} = \frac{\pi\sin(2\pi)}{2}+\frac{\cos(2\pi)}{4}- \frac{0\sin(2\cdot 0)}{2}+\frac{\cos(2\cdot 0)}{4}=0[/tex]
[tex] \int_{0}^{\pi}x\sin(x) = \left.-x\cos(x)+\sen(x)\right |_{0}^{\pi} = -\pi\cos(\pi)+\sen(\pi)- (-0\cdot \cos(\pi)+\sin(0)) = \pi[/tex]
Then, we have that
[tex]\int_{0}^{\pi} 2x(1-\cos(2x))-6x\sin(x)dx = \left.x^2\right|_{0}^{\pi} -2I_1-6I_2 = \pi^2-2\cdot 0 -6\pi = \pi(\pi-6)[/tex]
