Summer 2017 – Q.10

10. Verify Greens’ theorem in the plane for $latex \int_C\left(3x^2-8y^2\right)dx+\left(4y-6xy\right)dy$ where C is the boundary of the region defined by $latex y=\sqrt x,\;y=x^2$. [7M]

Given that $latex y=\sqrt x$

on squaring both sides

$latex \begin{array}{l}y^2=x\;and\\\\y=x^2\end{array}$ are the two parabolas intersecting at (0, 0) and A(1, 1)

Here,

$latex M=3x^2-8y^2$

differentiate w.r to y

$latex \frac{\partial M}{\partial y}=-16y$

Similarly

$latex N=4y-6xy$

differentiate w.r to x

$latex \frac{\partial N}{\partial x}=-6y$

$latex \begin{array}{l}\therefore\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=-6y+16\\\\\;\;\;\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=10y\end{array}$

By green’s theorem, we have

$latex \int\limits_c\left(Mdx+Ndy\right)=\int\limits_R\int\left[\frac{\displaystyle\partial N}{\displaystyle\partial x}-\frac{\displaystyle\partial M}{\displaystyle\partial y}\right]dxdy$

$latex =\int_{x=0}^{1}\int_{y=x^{2}}^{\sqrt{x}}10y dxdy$

$latex =\int_{x=0}^{1}\left [\int_{y=x^{2}}^{\sqrt{x}}10y dy \right ] dx$

$latex =\int_{x=0}^{1}\left [10 \frac{y^{2}}{2} \right ]_{x^{2}}^{\sqrt{x}} dx$

$latex \begin{array}{l}=\int_0^15\left[x-x^4\right]dx\\\\=5\int_0^1\left(x-x^4\right)dx\\\\=5\left[\frac{x^2}2-\frac{x^5}5\right]_0^1\\\\=5\left[\frac12-\frac15\right]\end{array}$

$latex \begin{array}{l}\therefore\int\limits_cMdx+Ndy\;=5\left(\frac12-\frac15\right)\\\\=5\left(\frac{5-2}{10}\right)\\\\=\frac32\\\\\therefore\int\limits_cMdx+Ndy\;=\frac32\end{array}$