Use Green’s Theorem to evaluate
∮C(x − 2y2) dx + (y4 + 2xy) dy where C consists of the line segment
from (0, 2) to (0, 4), followed by the curve with parametric equations x = 4 cos t, y = 4 sin t from (0, 4) to (−2, 2√3), then the line segment from (−2, 2√3) to (−1, √3), and finally the curve with parametric equations x = 2 sin t, y = 2 cos t from (−1, √3) to (0, 2).
Green's Theorem:
"\\oint_C (Ldx+Mdy)=\\iint_D(\\frac{\\partial M}{\\partial x}-\\frac{\\partial L}{\\partial y})dxdy"
"\\oint_C((x \u2212 2y^2) dx + (y^4 + 2xy) dy)=""\\iint_D(2y+4y)dxdy=\\iint_D6ydxdy"
Let's move to polar coordinates:
"x=r\\cos \\phi"
"y=r\\sin\\phi"
"2\\leq r\\leq 4"
"\\frac{\\pi}{2}\\leq\\phi\\leq \\arctan(\\frac{2\\sqrt 3}{-2})=\\frac{2\\pi}{3}"
"\\displaystyle\\iint_D6ydxdy=\\int_{\\frac{\\pi}{2}}^{\\frac{2\\pi}{3}}d\\phi\\int_2^46r\\sin\\phi \\cdot rdr=""\\int_{\\frac{\\pi}{2}}^{\\frac{2\\pi}{3}}\\sin\\phi d\\phi\\int_2^46r^2dr=-(\\cos\\phi|_{\\frac{\\pi}{2}}^{\\frac{2\\pi}{3}})\\cdot(2r^3|_2^4)=\\\\""=-(0.5-0)\\cdot2(4^3-2^3)=-56"
Answer: "\\oint_C((x \u2212 2y^2) dx + (y^4 + 2xy) dy)=-56" .
Comments
Leave a comment