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Use Green's Theorem to evaluate the following line integral. Assume the curve is oriented counterclockwise
The circulation line integral of 
F=(:3xy^(2),3x^(3)+y:) where 
C is the boundary of 
{(x,y):0 <= y <= sin x,0 <= x <= pi}

Use Green's Theorem to evaluate the following line integral. Assume the curve is oriented counterclockwise\newlineThe circulation line integral of F=3xy2,3x3+y F=\left\langle 3 x y^{2}, 3 x^{3}+y\right\rangle where C C is the boundary of {(x,y):0ysinx,0xπ} \{(x, y): 0 \leq y \leq \sin x, 0 \leq x \leq \pi\}

Full solution

Q. Use Green's Theorem to evaluate the following line integral. Assume the curve is oriented counterclockwise\newlineThe circulation line integral of F=3xy2,3x3+y F=\left\langle 3 x y^{2}, 3 x^{3}+y\right\rangle where C C is the boundary of {(x,y):0ysinx,0xπ} \{(x, y): 0 \leq y \leq \sin x, 0 \leq x \leq \pi\}
  1. Identify Vector Field and Region: Identify the vector field and the region over which the integral is to be evaluated. F=(3xy2,3x3+y)F = (3xy^2, 3x^3 + y) Region R: 0ysin(x)0 \leq y \leq \sin(x), 0xπ0 \leq x \leq \pi
  2. Write Green's Theorem: Write down Green's Theorem.\newlineGreen's Theorem relates a line integral around a simple, closed, positively oriented curve CC to a double integral over the region DD enclosed by CC:\newlineCFdr=D(QxPy)dA\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA\newlinewhere F=(P,Q)\mathbf{F} = (P, Q)
  3. Calculate Partial Derivatives: Calculate partial derivatives needed for Green's Theorem.\newlineP=3xy2P = 3xy^2, Q=3x3+yQ = 3x^3 + y\newlinedPdy=6xy\frac{dP}{dy} = 6xy, dQdx=9x2\frac{dQ}{dx} = 9x^2
  4. Set Up Double Integral: Set up the double integral using the partial derivatives. D(9x26xy)dA\iint_D (9x^2 - 6xy) \, dA
  5. Express Limits in Terms of x and y: Express the double integral in terms of x and y limits.\newlinex=0x=πy=0y=sin(x)(9x26xy)dydx\int_{x=0}^{x=\pi} \int_{y=0}^{y=\sin(x)} (9x^2 - 6xy) \, dy \, dx
  6. Integrate with Respect to y: Integrate with respect to y first.\newliney=0y=sin(x)(9x26xy)dy=[9x2y3xy2]y=0y=sin(x)\int_{y=0}^{y=\sin(x)} (9x^2 - 6xy) dy = [9x^2y - 3xy^2]_{y=0}^{y=\sin(x)}\newline=9x2sin(x)3xsin2(x)= 9x^2\sin(x) - 3x\sin^2(x)
  7. Integrate with Respect to x: Integrate with respect to xx. \newlinex=0x=π(9x2sin(x)3xsin2(x))dx\int_{x=0}^{x=\pi} (9x^2\sin(x) - 3x\sin^2(x)) \, dx\newlineThis integral needs to be evaluated, typically using integration techniques or numerical methods.

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