Vector Calculus - 1 - Mathematics MCQ

We have, 


So, the direction derivative in the direction of  

where, is an outward drawn unit normal vector to S.

So, from vector identity

Taking projection on three planes, we note that the surface S consists of three triangles, Δ OAB in XT- plane, Δ OBC in TZ-plane and Δ OAC in XZ-plane. Using two point formula, the equation of the line AB, BC, CA are respectively 3x + 2y = 6 , 2y + z = 6 , 3x + z = 6




So, by Stake’s theorem

So, potential function of

So, line integral of the vector from point (0, 0, 0) to (1, 1, 1) is

Cross checking from option (a),


which is correct answer.

We take,


= 1 - 2 + 1 = 0 
So, B is normal to A.

The cross-vector product of the vector always equals the vector. Perpendicular is the line and that will make the angle of 900with one another line. Therefore, when two given vectors are perpendicular then their cross product is not zero but the dot product is zero.

Correct Answer :- b

Explanation : Since, C is the curve y = ³from (0, 0) to (2, 8) 

So, let x = t ⇒ y = t³

If r is the position vector of any point on C, then

r(t) = xi + yj = ti +t³j

dr/dt = i + 3t²j 

F = (t² - t⁶)i + t⁴j

At (0, 0) ⇒ t = x = 0 and at (2, 8) ⇒ t = 2 

∫F.dr = ∫C(F.dr/dt)dt

∫(0 to 2) [(t² - t⁶)i + t⁴j] . (i + 3t²j)dt

∫(0 to 2) [(t² - 2t⁶)dt]

= 824/21

We have, curl 

Also we note that z coordinate of each vertex of the triangle is zero.
or The triangle lies in the xy-plane. So, 
So, curl 
In the figure, we have only considered the xy-plane. So, by Stoke’s Theorem

Since, 

Therefore,

The region R is bounded by two closed circles C₁ and C₂, so it is doubly connected. To apply it in Green’s theorem, we need to convert it into simply connected region. For it, we apply cut AD and consider the region R having simple closed curve ABCADEFDA in the anticlockwise direction. So, the directions shown in figure, (c) is correct option.

Since, 
Putting 

Given, 
Now, for divergence


Hence, vector field is divergence-free.
Now, for irrotational


 

Let V be the volume enclosed by the closed surface S, i.e., the volume in the first octant bounded by the cylinder y² I z² = 9 and the planes x = 0, x = 2. Then by Gauss’s divergence theorem, we have

Since, 
So, 
At (1, 3),

So,