The curves [tex]C_1,C_2,C_3[/tex] can be described parametrically by
[tex]\vec r_1(t)=7t\,\vec\imath,\quad0\le t\le1[/tex]
[tex]\vec r_2(t)=7\cos t\,\vec\imath+7\sin t\,\vec\jmath,\quad0\le t\le\dfrac\pi2[/tex]
[tex]\vec r_3(t)=7(1-t)\,\vec\jmath,\quad0\le t\le1[/tex]
so that the line integral over each component curve is
[tex]\displaystyle\int_{C_1}x^2y\,\mathrm dx+y^2x\,\mathrm dy=\int_0^1(0^2(7t)(7)+(7t)^2(0)(0))\,\mathrm dt=0[/tex]
[tex]\displaystyle\int_{C_2}x^2y\,\mathrm dx+y^2x\,\mathrm dy=\int_0^{\pi/2}((7\sin t)^2(7\cos t)(-7\sin t)+(7\cos t)^2(7\sin t)(7\cos t))\,\mathrm dt[/tex]
[tex]=\displaystyle7^4\int_0^{\pi/2}(\cos^2t\sin^2t-\cos^2t\sin^2t)\,\mathrm dt=0[/tex]
[tex]\displaystyle\int_{C_3}x^2y\,\mathrm dx+y^2x\,\mathrm dy=\int_0^1(0^2(7(1-t))(0)+(7(1-t))^2(0)(-7))\,\mathrm dt=0[/tex]
So the line integral over all of [tex]C[/tex] is 0.
We verify this result with Green's theorem, which says
[tex]\displaystyle\int_Cx^2y\,\mathrm dx+y^2\,\mathrm dx=\iint_D\frac{\partial(y^2x)}{\partial x}-\frac{\partial(x^2y)}{\partial y}\,\mathrm dx\,\mathrm dy=\iint_D(y^2-x^2)\,\mathrm dx\,\mathrm dy[/tex]
where [tex]D[/tex] is the region bounded by [tex]C[/tex]. In polar coordinates, this integral is
[tex]\displaystyle\int_0^{\pi/2}\int_0^7((r\sin\theta)^2-(r\cos\theta)^2)r\,\mathrm dr\,\mathrm d\theta=\left(-\int_0^{\pi/2}\cos2\theta\,\mathrm d\theta\right)\left(\int_0^7r^3\,\mathrm dr\right)=0[/tex]
