Question

* A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of $(x+y{)}^2$ on the path AB traversed in a counter-clocwise sense is

• A. $\frac{\pi }{2}-1$
• B. $\frac{\pi }{2}+1$
• C. $\frac{\pi }{2}$
• D. 1

Answer 1

The correct option is B $\frac{\pi }{2}+1$

Method I :
$x^2+y^2$ = 1
Let x = cos $\theta$,
y = sin$\theta$, $\theta$$ϵ[0,\frac{\pi }{2}]$
F = $(x+y{)}^2$ and dr =rd$\theta$
Now, ${\int }_{c}Fdr=\int {(x+y)}^{2}rd\theta ={\int }_0^{\frac{\pi }{2}}{(cos\theta +sin\theta )}^{2}d\theta$
$={\int }_0^{\frac{\pi }{2}}(1+sin2\theta )d\theta$
$={[\theta -\frac{cos2\theta }{2}]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}+1$

Method II :
We have equation for path AB
$x^2+y^2$ = 1
Differentiating above equation w.r.t. y, we get
$2x\frac{dx}{dy}+2y=0$

$\frac{dx}{dy}=-\frac{y}{x}$
Let $\Phi =(x+y{)}^2$. Then
${\int }_{AB}(x+y{)}^{2}dr={\int }_{AB}(x+y{)}^2\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}$
$={\int }_{AB}(x^2+y^2+2xy)\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy$
$={\int }_{AB}(x^2+y^2+2xy)\sqrt{(1+\frac{y^2}{x^2})}dy$
$={\int }_{AB}(1+2xy).\frac{1}{x}.dy$
$={\int }_{AB}(\frac{1}{x}+2y)dy$
$={\int }_{AB}(\frac{1}{\sqrt{1-y^2}}+2y)dy$
$={[si{n}^{-1}y+y^2]}_0^1$
$=\frac{\pi }{2}+1$