Winter 2016 - Q.10 - Grad Plus

# Winter 2016 – Q.10

10. Use Green’s theorem in the plane, evaluate $latex \int_c\left[\left(3x^2-8y^2\right)dx+\left(4y-6xy\right)dy\right]$ Where C is the boundary of the region bounded by $latex y=\sqrt x\;and\;y=\sqrt x$ [7M]

We have,

$latex y=\sqrt x,\;and\;y=x^2$

on squaring

$latex \therefore y^2=x,\;and\;y=x^2$ are two parabolas intersecting at 0 (0, 0) and A (1, 1).

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

Differentiate w r to y

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

and N = 4y – 6xy

differentiate w r to x

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

$latex \begin{array}{l}\therefore\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=-6y-\left(-16y\right)\\\\=10y\end{array}$

By greens theorem
we have,

$latex \int_c\left(M\;dx+N\;dy\right)$

$latex =\int_R\int\left[\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right]dx\;dy$

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

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

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

$latex =5\int_{0}^{1}\left [y^{2} \right ]_{x^{2}}^{\sqrt{x}} dx$

$latex \begin{array}{l}=5\int_0^1\left[x-x^4\right]dx\\\\=5\left[\frac{x^2}2-\frac{x^5}5\right]_0^1\\\\=5\left[\frac12-\frac15\right]\\\\=5\left(\frac3{10}\right)\\\\=\frac32\end{array}$

$latex \therefore\int_c\left(M\;dx+N\;dy\right)=\frac32$

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