(b) Using Green's int c x^ 2 ydx + x ^ 2 * dxy theorem, evaluate where C is the triangle
joining the points (0, 0), (1, 0) and (0, 1).

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(b) Using Green&#039;s int c x^ 2 ydx + x ^ 2 * dxy theorem, evaluate where C is the trianglejoining the points (0, 0), (1, 0) and (0, 1).&amp;nbsp;

Don Sumner · Accepted Answer

To evaluate the line integral ∮Cx2ydx+x2dy using Green&#039;s theorem, we need to find the vector field 𝐅(x,y) such that 𝐅=P𝐢+Q𝐣, where P=x2y and Q=x2.By applying Green&#039;s theorem, the line integral can be rewritten as a double integral over the region enclosed by the curve C:∮Cx2ydx+x2dy=∬R(∂Q∂x−∂P∂y)dAwhere ∂Q∂x represents the partial derivative of Q with respect to x, and ∂P∂y represents the partial derivative of P with respect to y.Let&#039;s calculate these partial derivatives:∂Q∂x=∂∂x(x2)=2x∂P∂y=∂∂y(x2y)=x2Substituting these values into the integral expression, we have:∬R(2x−x2)dANow, we need to determine the region R enclosed by the triangle joining the points (0, 0), (1, 0), and (0, 1).The region R is a right triangle with vertices (0, 0), (1, 0), and (0, 1). We can evaluate the double integral by integrating over this triangular region.Using the limits of integration, we have:∬R(2x−x2)dA=∫01∫0x(2x−x2)dydx+∫10∫01−x(2x−x2)dydxSimplifying the integrals, we obtain:∫01∫0x(2x−x2)dydx+∫10∫01−x(2x−x2)dydx=∫01xy−x2y2|0x+∫01xy−x2y2|1−x0dxEvaluating the integrals, we get:∫01xy−x2y2|0x+∫01xy−x2y2|1−x0dx=∫01x2−x32dx+∫01(1−x)y−(1−x)2y2dxSimplifying further, we have:∫01x2−x32dx+∫01(1−x)y−(1−x)2y2dx=[x33−x48]01+[y2x2−y3x3+y6x4]01Evaluating the limits of integration, we obtain:[x33−x48]01+[y2x2−y3x3+y6x4]01=13−18+y2−y3+y6Simplifying the expression, we have:13−18+y2−y3+y6=1324+y4Therefore, the value of the line integral ∮Cx2ydx+x2dy along the triangle C joining the points (0, 0), (1, 0), and (0, 1) is 1324+y4.

(b) Using Green's int c x^ 2 ydx

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