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I found that the explanation to option (d) is wrong as curl of vector $F$ is not zero hence not irrotational. What say?

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2 Answers 2

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You probably made a mistake in your calculations.

$$\mbox{curl}\, \vec F = \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 4y^2+\frac{3x^2y}{z^2} & 8xy+\frac{x^3}{z^2} & 11-\frac{2x^3y}{z^3} \end{vmatrix} \\[38pt] = \left( \frac{\partial}{\partial y} \left(11-\frac{2x^3y}{z^3} \right) - \frac{\partial}{\partial z} \left( 8xy+\frac{x^3}{z^2} \right)\right)\hat i \\[18pt] + \left( \frac{\partial}{\partial z} \left( 4y^2+\frac{3x^2y}{z^2} \right) - \frac{\partial}{\partial x} \left( 11-\frac{2x^3y}{z^3} \right)\right) \hat j \\[18pt] + \left( \frac{\partial}{\partial x} \left( 8xy+\frac{x^3}{z^2} \right) - \frac{\partial}{\partial y} \left( 4y^2+\frac{3x^2y}{z^2} \right) \right) \hat k \\[18pt] = \cdots = 0. $$

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    Got it. A sign mistake – 2017-01-18
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    Those are the most dangerous :-). – 2017-01-18
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    @StackTD is the second term supposed to be minus? It's still zero but just double checking. – 2017-01-18
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    There's either a minus, or you switch the order of the two derivatives (which is what I did; it comes down to getting the minus inside the brackets). – 2017-01-18
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You are wrong. On finding curl F using determinat we got 0. Hence irrotational.

Curl F = $\left(- \frac{2x^3}{z^3} + \frac{2x^3}{z^3} \right) \hat{i} - \left(\frac{-6x^2y}{z^3} - \frac{-6x^2y}{z^3} \right) \hat{j} + \left(8y + \frac{3x^2}{z^2} - 8y - \frac{3x^2}{z^2} \right) \hat{k}$

= 0

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    Sorry few terms mistakes before as I am solving it on my mobile. – 2017-01-18