As shown in the figure, C is the arc from the point (3,0) to the point (0,3) on the circle $x^{2} + y^{2} = 9$ . The value of the integral $\int_{c} (y^{2} + 2 y x) d x + (2 x y + x^{2}) d y$ is _______(up to 2 decimal places)

0

Open in App

Solution

The correct option is A 0
I= $\int_{c} [(y^{2} + 2 x y) d x + (2 x y + x^{2} d y]$
= $\int_{c} (f_{1} d x + f_{2} d y)$
On comparison
$\to f = f_{1}^i + f_{2}^j + f_{3}^k = (y^{2} + 2 x y)^i + (x^{2} + 2 x y)^j$
Curl $\to f$ is conservative, i.e path independent, so we can integrate along easiest possible path 1.e. along the straight line A(3,0) to B(0.3).
AB: $\frac{x}{3} + \frac{y}{3} = 1 \Rightarrow y = 3 - x$
so dy = -dx, $3 \leq x \leq 0$ .
I = $\int_{c} [(y^{2} + 2 x y) d x + (x^{2} + 2 x y)]$
= $\int_{A B} [(3 - x)^{2} + 2 x (3 - x) d x] + [x^{2} + 2 x (3 - x) (- d x)]$
= $\int_{x = 3}^{0} (9 - 6 x) d x = (9 x - 3 x^{2})_{3}^{0} = 0$

Suggest Corrections

Similar questions

x^{2} + 4 x + (y - 3)^{2} = 0

x^{2} + y^{2} = 9

x^{2} + y^{2} = 4

x^{2} + y^{2} = 9

x^{2} + y^{2} = 9

π

3. 14159265358979323846264338327950288

0

9

x^{2} + y^{2} = 9

(\frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}})

\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1

x^{2} + y^{2} = 4

Join BYJU'S Learning Program