We know that curl for any vector quantity is given by
\(\bigtriangledown.\vec{r}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\y^2 z^3&x^2 z^2&(x-2y)\end{bmatrix}\)
\(\bigtriangledown.\vec{r}=\vec{i}\left (\frac{\partial(x-2y)}{\partial y}-\frac{\partial(x^2 z^2)}{/\partial z}\right )-\vec{j}\left (\frac{\partial(x-2y)}{\partial x}-\frac{\partial(y^2 z^3)}{\partial z}\right )+\vec{k}\left(\frac{\partial(x^2 z^2)}{\partial x}-\frac{\partial(y^2 z^3)}{\partial y}\right )\) \(\bigtriangledown.\vec{r}=-2\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\).

Here insect will get warm quickly if it moves normal to given
Surface in space.
As \(T = x^2+y^2-z \)
\(? ? T = (\frac{?}{?x}\hat{i} + \frac{?}{?x}\hat{j} + \frac{?}{?x}\hat{k})(x^2+y^2-z) \)
\( = 2x\hat{i} + 2y\hat{j} – \hat{k} \)
Now \( ? T |_{(1,1,2)} = 2\hat{i} + 2\hat{j} – \hat{k} \)
? required direction\( = \frac{? T}{|? T|} \) \(=\frac{2\hat{i} + 2\hat{j} – \hat{k}}{\sqrt{(4+4+1)}} \)
\(= \frac{2\hat{i} +2\hat{j} – \hat{k}}{3} \)
\(= \frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3}. \)

P = x³ y and Q = xy³
\(\frac{?P}{?y} = x^3\) and \(\frac{?Q}{?x}= y^3 \)
Now since y aren?t equal, vector is not a conservative vector or field.

Let directional derivative is zero along \(a1\hat{i} + a2\hat{j}\)
\(Then ? f = \hat{i} \frac{?f}{?x} +\hat{j} \frac{?f}{?y} = \hat{i} (2xy+2y^2) + \hat{j}(x^2+4xy) \)
\(i.e.[? f]_{(1,3)} = 24\hat{i} + 13\hat{j} \)
\([? f]_{(1,3)} . (a_1\hat{i} +a_2\hat{j}) = 0 \)
\((a_1\hat{i} +a_2\hat{j}) . (a_1\hat{i} + a_2\hat{j}) = 0 \)
\(24a_1 + 13a_2 = 0 ? \frac{a1}{13}=\frac{(-a2)}{24} \)
? Direction are \(\overline{u1} = 13\hat{i} ? 24\hat{j}, \overline{u2} = -13\hat{i} + 24\hat{j}. \)

First find gradient of W which is (2xy²+z) a_x + 2yx² a_y + x a_z
At (1,2,0) gradient of W is 8 a_x + 4 a_y + 1 a_z
\(\frac{dW}{dl} = \) (Gradient of W ) . al
\( = (8,4,1) . \frac{(3,4,6)}{\sqrt{(9+16+36)}} \)
\(=5.88897 = 6.\)

The formula to convert any vector from Spherical coordinates to Cartesian coordinates is given by \(\begin{bmatrix}
Px\\
Py\\
Pz\\
\end{bmatrix} \) \(= \begin{bmatrix}
sin\ta cos?? & cos\ta cos?? & -sin?\\
sin\ta sin?? & cos?\ta sin? & cos?\\
cos\ta & -sin\ta & 0\\
\end{bmatrix} \) \(= \begin{bmatrix}
r\\
0\\
cos\ta\\
\end{bmatrix} \)
after substituting values of vector. Now, solving matrix we get answer
\(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \).

\(W = ?zcos(?) \)
Gradient of \(W = \frac{1}{?} \frac{?W}{??} a_? + \frac{1}{?} \frac{?W}{??}a_y + \frac{?W}{?z}a_z \)
\(= zcos(?)a_? – z sin(?) a_y + ?cos(?) a_z \)

We know that divergence of a vector is given by
\(\bigtriangledown.\vec{F}=\frac{\partial(x^2 y)}{\partial x}+\frac{\partial(3x+y)}{\partial y}+\frac{\partial(y^3 z)}{\partial z}\)
\(\bigtriangledown.\vec{F}=y^3+2xy+1\)
We know that curl for any vector quantity is given by
\(\bigtriangledown.\vec{r}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2 y&3x+y&(y^3 z)\end{bmatrix}\)
\(\bigtriangledown.\vec{r}=\vec{i}\left (\frac{\partial(y^3 z)}{\partial y}-\frac{\partial(3x+y)}{\partial z}\right )-\vec{j}\left (\frac{\partial(y^3 z)}{\partial x}-\frac{\partial(x^2 y)}{\partial z}\right )+\vec{k}\left (\frac{\partial(3x+y)}{\partial x}-\frac{\partial(x^2 y)}{\partial y}\right )\)
\(\bigtriangledown.\vec{r}=\vec{i}(3y^2 z)+\vec{k}(3-x^2)\).

By definition: A vector field whose divergence comes out to be zero or
Vanishes is called as a Solenoidal Vector Field.
i.e.
If \(?. \vec{f} = 0 ↔ \vec{f} \) is a Solenoidal Vector field.

We first have to convert given points to Cartesian coordinates from Spherical coordinates. Then we have to apply formula for distance between two points in Cartesian coordinates and find distance.