- #1

WMDhamnekar

MHB

- 329

- 27

- Homework Statement:
- Compute ## \oint_C x^2z dx + 3xdy -y^3 dz ## where C is the unit circle ##x^2 +y^2 =1## oriented counter-clockwise

- Relevant Equations:
- ##\int_C f\cdot dr = \iint\limits_\Sigma curl(f)\cdot n d\sigma##

##curl([x^2z, 3x , -y^3],[x,y,z]) =[-3y^2 ,x^2,3]##

The unit normal vector to the surface ##z(x,y)=x^2+y^2## is ##n= \frac{-2xi -2yj +k}{\sqrt{1+4x^2 +4y^2}}##

##[-3y^2,x^2,3]\cdot n= \frac{-6x^2y +6xy^2}{\sqrt{1+4x^2 + 4y^2}}##

Since ##\Sigma## can be parametrized as ##r(x,y) = xi + yj +(x^2 +y^2)k## for (x,y) in the region ##D = \{(x,y): x^2 + y^2 \leq 1\}## then,

##\begin{align*}\iint\limits_{\Sigma} (curl f) \cdot n d\sigma &= \iint\limits_{D} curl f \cdot n \parallel \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}\parallel dA\\

&=\iint\limits_{D} 3 - 2x^2y + 6xy^2 dA\\

&= \int_0^{2*\pi}\int_0^1 (3 - 2r^3 \cos^2{(\theta)}\sin{(\theta)} + 6r^3\cos{(\theta)}\sin^2{(\theta)})r dr d\theta \\

&= 3\pi

\end {align*}##

But the answer provided by author is ##3\pi## How is that? Where I am wrong?

The unit normal vector to the surface ##z(x,y)=x^2+y^2## is ##n= \frac{-2xi -2yj +k}{\sqrt{1+4x^2 +4y^2}}##

##[-3y^2,x^2,3]\cdot n= \frac{-6x^2y +6xy^2}{\sqrt{1+4x^2 + 4y^2}}##

Since ##\Sigma## can be parametrized as ##r(x,y) = xi + yj +(x^2 +y^2)k## for (x,y) in the region ##D = \{(x,y): x^2 + y^2 \leq 1\}## then,

##\begin{align*}\iint\limits_{\Sigma} (curl f) \cdot n d\sigma &= \iint\limits_{D} curl f \cdot n \parallel \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}\parallel dA\\

&=\iint\limits_{D} 3 - 2x^2y + 6xy^2 dA\\

&= \int_0^{2*\pi}\int_0^1 (3 - 2r^3 \cos^2{(\theta)}\sin{(\theta)} + 6r^3\cos{(\theta)}\sin^2{(\theta)})r dr d\theta \\

&= 3\pi

\end {align*}##

But the answer provided by author is ##3\pi## How is that? Where I am wrong?

Last edited: