Question

If \[ I=\int_C x^2y\,dx-xy^2\,dy, \] where \(C\) is the circle \[ x^2+y^2=16, \] then \(2I=\)

• A. \(-128\pi\)
• B. \(-256\pi\)
• C. \(-512\pi\)
• D. \(0\)

Hint:

For a closed curve, \[ \boxed{ \oint_C M\,dx+N\,dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dA } \] Use polar coordinates whenever the region is a circle.

Answer 1

The Correct Option is B

Step 1: Apply Green's theorem. For a closed curve, \[ \oint_C M\,dx+N\,dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dA. \] Here, \[ M=x^2y, \] \[ N=-xy^2. \]

Step 2: Find the required partial derivatives. \[ \frac{\partial N}{\partial x} = -y^2, \] \[ \frac{\partial M}{\partial y} = x^2. \] Hence, \[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -(x^2+y^2). \] Therefore, \[ I = -\iint_R (x^2+y^2)\,dA. \]

Step 3: Convert into polar coordinates. Using \[ x^2+y^2=r^2, \] and \[ dA=r\,dr\,d\theta, \] we get \[ I = - \int_0^{2\pi} \int_0^4 r^3\,dr\,d\theta. \] \[ = - \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_0^4 d\theta. \] \[ = - \int_0^{2\pi} 64\,d\theta. \] \[ = -128\pi. \] Hence, \[ 2I = 2(-128\pi) = -256\pi. \] Therefore, \[ \boxed{-256\pi} \] is the correct answer. Thus, \[ \boxed{(B)} \] is the correct answer.