n this and the following problem you will consider the integral ∮C8ysin(8x)dx+3xydy on the closed curve C consisting of the line segments from (0,0) to (3,1) to (0,1) to (0,0). Here, you evaluate the line integral along each of these segments separately (as you would have before having attained a penetrating and insightful knowledge of Green's Theorem), and in the following problem you will apply Green's Theorem to find the same integral. Note that you can check your answers between the two problems, because the value of the final integral will be the same (that is, the sum you find below must be equal to the final anser in the following problem). Evaluate the integral above by finding the integral from (0,0) to (3,1), adding the integral from (3,1) to (0,1), and adding the integral from (0,1) to (0,0): ∮C8ysin(8x)dx+3xydy=

Without Green's theorem:Parameterize each leg of this triangle by[tex]\vec r_1(t)=(1-t)(0,0)+t(3,1)=(3t,t)[/tex][tex]\vec r_2(t)=(1-t)(3,1)+t(0,1)=(3-3t,1)[/tex][tex]\vec r_3(t)=(1-t)(0,1)+t(0,0)=(0,1-t)[/tex]each with [tex]0\le t\le1[/tex].Then the line integral[tex]\displaystyle\int_C(8y\sin8x\,\mathrm dx+3xy\,\mathrm dy)[/tex]computed over each leg is[tex]\displaystyle\int_0^1(24t\sin24t+9t^2)\,\mathrm dt=3-\cos24+\frac{\sin24}{24}[/tex][tex]\displaystyle\int_0^1-24\sin(24-24t)\,\mathrm dt=-2\sin^212[/tex][tex]\displaystyle\int_0^10\,\mathrm dt=0[/tex]so that the total value of the line integral is[tex]\displaystyle\int_C(8y\sin8x\,\mathrm dx+3xy\,\mathrm dy)=3-\cos24-2\sin^212+\frac{\sin24}{24}\approx1.96[/tex]With Green's theorem:We have[tex]\displaystyle\int_C(8y\sin8x\,\mathrm dx+3xy\,\mathrm dy)=\iint_D\left(\frac{\partial(3xy)}{\partial x}-\frac{\partial(8y\sin8x)}{\partial y}\right)\,\mathrm dA[/tex][tex]\displaystyle=\iint_D(3y-8\sin8x)\,\mathrm dA[/tex]where [tex]D[/tex] denotes the interior of [tex]C[/tex]. This is equivalent to[tex]\displaystyle\int_0^1\int_0^{3y}(3y-8\sin8x)\,\mathrm dx\,\mathrm dy=2+\frac{\sin24}{24}\approx1.96[/tex]