For a scalar function (x, y, z) = x2 + 3y2 + 2z2, the directional derivative at the point P( 1, 2, -1) is the direction of a vector is
So, the direction derivative in the direction of
Use the divergence theorem the value of where, S is any closed surface enclosing volume V.
where, is an outward drawn unit normal vector to S.
Find the value of
So, from vector identity
If then the value of div at the point (1, 1, -1) will be
Apply Stoke’s theorem, the value of where C is the boundary of the triangle with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6) is
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
The line integral from the origin to the point P( 1,1,1) is
So, potential function of
So, line integral of the vector from point (0, 0, 0) to (1, 1, 1) is
If are to arbitrary vectors with magnitudes a and b respectively, will be equal to
Cross checking from option (a),
which is correct answer.
Divergence of the three-dimensional radial vector field is
Which of the following holds for any non-zero vector
A vector normal to is
= 1 - 2 + 1 = 0
So, B is normal to A.
If and curve C is the arc of the curve y = x3 from (0, 0 ) to (2, 8), then the value of
Since, C is the curve y = x3 from (0, 0) to (2, 8)
So, let x = t ⇒ y = t3
If is the position vector of any point on C, then
At (0, 0) ⇒ t = x = 0 and at (2, 8) ⇒ t = 2
The value of by Stoke’s theorem, where and C is the boundary of the triangle with vertices at ( 0 ,0 , 0 ) , ( 1 , 0 , 0 ) and ( 1 ,1 , 0 ) is
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,
In the figure, we have only considered the xy-plane. So, by Stoke’s Theorem
If is the reciprocal system to the vectors then the value of is
Scalar triple product is equal to
R is a closed planar region as shown by the shaded area in the figure below. Its boundary C consists of the circles C1 and C2.
If are all continuous everywhere in R, Green’s theorem states that
Which one of the following alternatives correctly depicts the direction of integration along C?
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.
The value of is
The vector field are unit vectors) is
Now, for divergence
Hence, vector field is divergence-free.
Now, for irrotational
Use Gauss’s divergence theorem to find where and S is the closed surface in the first octant bounded by y2 + z2 = 9 and x = 2.
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
For the scalar field magnitude of the gradient at the point (1, 3) is
At (1, 3),
is equal to