# Problems on Line Integral-3

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Example 1 :

A particle moves counterclockwise along the curve $3x^2 + y^2 = 3$ from (1, 0) to a point P, under the action of the force

$\vec F\left( {x,y} \right) = \dfrac{x}{y}\hat i + \dfrac{y}{x}\hat j.$

Prove that there are two possible locations of P such that the work done by $\vec F$ is 1.

Solution :

$\dfrac{{{x^2}}}{1} + \dfrac{{{y^2}}}{3} = 1$

Point on ellipse is represented as

$(\cos \theta ,\sqrt 3 \sin \theta )$

$\int {\vec F \cdot d\vec r = \int {\left( {\dfrac{x}{y}\hat i + \dfrac{y}{x}\hat j} \right) \cdot (dx\hat i + dy\hat j)} }$

$\int {\vec F \cdot d\vec r = \int {\left( {\dfrac{x}{y}\hat i + \dfrac{y}{x}\hat j} \right) \cdot (dx\hat i + dy\hat j)} }$

$= \int {\dfrac{x}{y}dx + \dfrac{y}{x}dy}$

$= \int_0^\theta {\dfrac{{\cos \theta }}{{\sqrt 3 \sin \theta }}( - \sin \theta )d\theta } + \dfrac{{\sqrt 3 \sin \theta }}{{\cos \theta }} \cdot \sqrt 3 \cos \theta d\theta$

$= \int_0^\theta {\left( { - \dfrac{1}{{\sqrt 3 }}\cos \theta + 3\sin \theta } \right)d\theta }$

$= \left. { - \dfrac{1}{{\sqrt 3 }}\sin \theta - 3\cos \theta } \right|_0^\theta$

$= - \dfrac{1}{{\sqrt 3 }}\sin \theta - 3\cos \theta + 3$

Work done in equal to 1

So, ${1}{{\sqrt 3 }}\sin \theta - 3\cos \theta + 3 = 1$

$\Rightarrow \ \ \ \ \ \ \ \ \dfrac{1}{{\sqrt 3 }}\sin \theta + 3\cos \theta = 2$

$\Rightarrow \ \ \ \ \ \ \ \ {\left( {\dfrac{1}{{\sqrt 3 }}\sin \theta } \right)^2} = {(2 - 3\cos \theta )^2} \$

$\Rightarrow \ \ \ \ \ \ \ \ \dfrac{1}{3}{\sin ^2}\theta = 4 + 9{\cos ^2}\theta - 12\cos \theta \$

$\Rightarrow \ \ \ \ \ \ \ \ 28 \cos^2 \theta - 36 \cos \theta + 11 = 0$

$\Rightarrow \ \ \ \ \ \ \ \ (2 \cos \theta - 1)(14 \cos \theta - 11) = 0$

$\cos \theta = \dfrac{1}{2},\dfrac{{11}}{{14}}$

So, there are two value of $\theta$ i.e., two possible location of P such that the work done by $\vec F$ is 1.

Example 2 :

Find the circulation of the field

$\vec F = -{x^2}y\hat i + x{y^2}\hat j + ({y^3}-{x^3})\hat k$

around the curve C, where C is the intersection of the sphere $x^2 + y^2 + z^2 = 25$ and the plane z = 3. The orientation of the curve C is counterclockwise when viewed from above.

Solution :

$\vec F = -{x^2}y\hat i + x{y^2}\hat j + ({y^3}-{x^3})\hat k$

C is the curve of intersection of surfaces

$x^2 + y^2 + z^2 = 25, z = 3$

So, $x^2 + y^2 = 16$

$\vec F \cdot d\vec r = x^2ydx + xy^2dy + (y^3 - x^3)dz$

For curve C, z = 3, dz = 0

$\oint\limits_C {\vec F \cdot d\vec r = \int_C { - {x^2}ydx + x{y^2}dy} }$

Let $x = 4 \cos \theta, y = 4 \sin \theta$

$\oint\limits_C {\vec F} \cdot \,\,d\vec r\,\, = \,\,\int\limits_0^{2\pi } {(256{{\cos }^2}\theta {{\sin }^2}\theta d\theta } + \,256{\cos ^2}\theta {\sin ^2}\theta )d\theta$

$=512 \int\limits_0^{2\pi } {{{\sin }^2}} \theta {\cos ^2}\theta d\,\theta$

$= 512 \times 4\int\limits_0^{\pi /2} {{{\sin }^2}} \theta {\cos ^2}\theta \,d\,\theta$

$= 2048 \dfrac{\left \lceil 3/2 \right. \left \lceil \dfrac{3}{2} \right.}{2 \left \lceil 3 \right.} = 128 \pi$