A surface $S (x, y) = 2 x + 5 y - 3$ is integrated once over a path consisting of the points that satissfy $(x + 1)^{2} + (y - 1)^{2} = \sqrt{2}$ . The integral evaluates to

17 \sqrt{2}

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\frac{17}{\sqrt{2}}

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\frac{\sqrt{2}}{17}

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0

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Solution

The correct option is D $0$
Equation of Circle:-
$(x + 1)^{2} + (y - 1)^{2} = \sqrt{2}$
Let $x + 1 = X \Rightarrow x = X - 1$
$y - 1 = Y \Rightarrow y = Y + 1$
$∴ X^{2} + Y^{2} = \sqrt{2}$
Let the parametric points on $X^{2} + Y^{2} = \sqrt{2}$
$X = r c o s θ$
$Y = r s i n θ$
Now, Equation of Surface:
$S (x, y) = 2 x + 5 y - 3$
$\Rightarrow = 2 (X - 1) + 5 (Y + 1) - 3$
$\Rightarrow = 2 X + 5 Y$
$∴ I = \int_{S} S (x, y) d x d y$
$= \int_{S} S (r, θ) r d r d θ$
$= \int_{S} [2 (r c o s θ) + 5 (r s i n θ)] r d r . d θ$
$= \int_{S} r (2 c o s θ + 5 s i n θ) . r d r . d θ$
$= \int_{S} (2 c o s θ + 5 s i n θ) r^{2} d r d θ$
$= \int_{0}^{r} r^{2} d r . \int_{θ = 0}^{2 π} (2 c o s θ + 5 s i n θ) d θ = 0$

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