Question

Use Green's Theorem to evaluate int_C vec(F)*d vec(r) where vec(F)(x,y)=xy^2 i + (1-xy^3)j and C is the parallelogram with vertices (-1,2), (-1,-1),(1,1)and(1,4). The orientation of C is counterclockwise.

Use Green's Theorem to evaluate \(\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}\) where \(\displaystyle\vec{{{F}}}{\left({x},{y}\right)}={x}{y}^{{2}}{i}+{\left({1}-{x}{y}^{{3}}\right)}{j}\) and C is the parallelogram with vertices (-1,2), (-1,-1),(1,1)and(1,4).
The orientation of C is counterclockwise.

Expert Answers (1)

2021-03-02
Step 1
Using Green theorem the integral becomes:
\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}=\int\int_{{S}}{\left(\frac{\partial}{{\partial{x}}}{\left({1}-{x}{y}^{{3}}\right)}-\frac{\partial}{{\partial{y}}}{x}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
where S is the surface enclosed by C.
Step 2
Thus, solve the integral noting that \(\displaystyle{x}\le{y}\le{x}+{3}{\quad\text{and}\quad}−{1}\le{x}\le{1}\):
\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}=\int\int_{{S}}{\left(\frac{\partial}{{\partial{x}}}{\left({1}-{x}{y}^{{3}}\right)}-\frac{\partial}{{\partial{y}}}{x}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle=\int\int_{{S}}{\left(-{y}^{{3}}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}{\int_{{{y}={x}}}^{{{x}+{3}}}}{\left(-{y}^{{3}}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}{\left(-\frac{{{\left({x}+{3}\right)}^{{4}}-{x}^{{4}}}}{{4}}-{x}{\left({\left({x}+{3}\right)}^{{2}}-{x}^{{2}}\right)}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}\frac{{-{12}{x}^{{3}}-{78}{x}^{{2}}-{144}{x}-{81}}}{{4}}{\left.{d}{x}\right.}\)
Step 3
Neglecting all odd powers of x:
\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}={\int_{{{x}=-{1}}}^{{1}}}\frac{{-{78}{x}^{{2}}-{81}}}{{4}}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{1}}{{4}}{\int_{{{x}=-{1}}}^{{1}}}-{78}{x}^{{2}}-{81}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{{1}}{{2}}{\left(-\frac{{78}}{{3}}-{81}\right)}\)
\(\displaystyle=-\frac{{107}}{{2}}\)
Step 4
Thus, the integral is \(\displaystyle-\frac{{107}}{{2}}=-{53.5}\).
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