Q:

Evaluate the integral ∮cf⋅dr for f=(3x2y)i+(2x2−3xy2)j on the curve c consisting of the x-axis from x=0 to x=2, the arc of the circle x2+y2=4 up to the line y=x, and the line y=xdown to the origin.

Accepted Solution

A:
Write the line integral as

[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{\mathcal C}P\,\mathrm dx+Q\,\mathrm dy[/tex]

In other words, [tex]\mathbf f(x,y)=P(x,y)\,\mathbf i+Q(x,y)\,\mathbf j[/tex], and [tex]\mathrm d\mathbf r=\mathrm dx\,\mathbf i+\mathrm dy\,\mathbf j[/tex].

The curve [tex]\mathcal C[/tex] is closed, so we can apply Green's theorem:

[tex]\displastyle\int_{\mathcal C}P\,\mathrm dx+Q\,\mathrm dy=\iint_{\mathcal D}\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{\mathcal D}\bigg((4x-3y^2)-(3x^2)\bigg)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{\mathcal D}(4x-3x^2-3y^2)\,\mathrm dx\,\mathrm dy[/tex]

Convert to polar coordinates to write this as

[tex]\displaystyle\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=2}(4r\cos\theta-3r^2\cos^2\theta-3r^2\sin^2\theta)r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_{\theta=0}^{\theta=\pi/4}\int_{r=0}^{r=2}(4r^2\cos\theta-3r^3)\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\dfrac{16\sqrt2}3-3\pi[/tex]