Answer to Question #104421 in Calculus for Ajay

Question

Use Green’s theorem, and apply it to evaluate ∫(3x^2–4y)dx–(2x+y^3)dy where C is the ellipse 4x^2+9y^2=25

Answer 1

ANSWER: $\frac{25π}{3}$

EXPLANATION:

By the Green's theorem $\quad \oint { (3{ x }^{ 2 }-4y)dx-(2x+{ y }^{ 3 } } )dy$ = $=\quad \underset { \quad }{ \underset { A }{ \iint } \left[ \frac { \partial \left( -2x-{ y }^{ 3 } \right) \quad \quad }{ \partial x } -\frac { \partial \left( 3{ x }^{ 2 }-4y \right) \quad \quad }{ \partial y } \quad \right] dxdy } \quad=$ $=\quad \underset { \quad }{ \underset { A }{ \iint } \left[ -2+4\quad \quad \right] dxdy } ,\quad \quad$ where A is the region bounded by ellipse ${ 4x }^{ 2\quad }+9{ y }^{ 2 }=25$ . The semi-axes of the ellipse are $\frac { 5 }{ 2 }$ and $\frac{5}{3}$ . The area of the ellipse is $\frac{25π}{6}$ . Therefore $\quad \underset { \quad }{ \underset { A }{ \iint } \left[ -2+4\quad \quad \right] dxdy } =2\left(\frac{25π}{6}\right)=\frac{25π}{3}$