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VECTOR CALCULUS 1.10 GRADIENT OF A SCALAR 1.11 DIVERGENCE OF A VECTOR 1.12 DIVERGENCE THEOREM 1.13 CURL OF A VECTOR 1.14 STOKES’S THEOREM 1.15 LAPLACIAN OF A SCALAR

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1.10 GRADIENT OF A SCALAR Suppose is the temperature at, and is the temperature at as shown.

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The differential distances are the components of the differential distance vector : However, from differential calculus, the differential temperature: GRADIENT OF A SCALAR (Cont’d)

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But, So, previous equation can be rewritten as: GRADIENT OF A SCALAR (Cont’d)

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The vector inside square brackets defines the change of temperature corresponding to a vector change in position. This vector is called Gradient of Scalar T. GRADIENT OF A SCALAR (Cont’d) For Cartesian coordinate:

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GRADIENT OF A SCALAR (Cont’d) For Circular cylindrical coordinate: For Spherical coordinate:

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EXAMPLE 10 Find gradient of these scalars: (a) (b) (c)

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SOLUTION TO EXAMPLE 10 (a) Use gradient for Cartesian coordinate:

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SOLUTION TO EXAMPLE 10 (Cont’d) (b) Use gradient for Circular cylindrical coordinate:

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SOLUTION TO EXAMPLE 10 (Cont’d) (c) Use gradient for Spherical coordinate:

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1.11 DIVERGENCE OF A VECTOR Illustration of the divergence of a vector field at point P: Positive Divergence Negative Divergence Zero Divergence

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DIVERGENCE OF A VECTOR (Cont’d) The divergence of A at a given point P is the outward flux per unit volume:

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DIVERGENCE OF A VECTOR (Cont’d) What is ?? Vector field A at closed surface S

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Where, And, v is volume enclosed by surface S DIVERGENCE OF A VECTOR (Cont’d)

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For Cartesian coordinate: For Circular cylindrical coordinate: DIVERGENCE OF A VECTOR (Cont’d)

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For Spherical coordinate: DIVERGENCE OF A VECTOR (Cont’d)

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EXAMPLE 11 Find divergence of these vectors: (a) (b) (c)

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18 (a) Use divergence for Cartesian coordinate: SOLUTION TO EXAMPLE 11

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(b) Use divergence for Circular cylindrical coordinate: SOLUTION TO EXAMPLE 11 (Cont’d)

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(c) Use divergence for Spherical coordinate:

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It states that the total outward flux of a vector field A at the closed surface S is the same as volume integral of divergence of A. 1.12 DIVERGENCE THEOREM

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EXAMPLE 12 A vector field exists in the region between two concentric cylindrical surfaces defined by ρ = 1 and ρ = 2, with both cylinders extending between z = 0 and z = 5. Verify the divergence theorem by evaluating: (a) (b)

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SOLUTION TO EXAMPLE 12 (a) For two concentric cylinder, the left side: Where,

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SOLUTION TO EXAMPLE 12 Cont’d)

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Therefore SOLUTION TO EXAMPLE 12 Cont’d)

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(b) For the right side of Divergence Theorem, evaluate divergence of D So,

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1.13 CURL OF A VECTOR The curl of vector A is an axial (rotational) vector whose magnitude is the maximum circulation of A per unit area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum.

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Where, CURL OF A VECTOR (Cont’d)

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The curl of the vector field is concerned with rotation of the vector field. Rotation can be used to measure the uniformity of the field, the more non uniform the field, the larger value of curl.

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For Cartesian coordinate: CURL OF A VECTOR (Cont’d)

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For Circular cylindrical coordinate: CURL OF A VECTOR (Cont’d)

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For Spherical coordinate:

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EXAMPLE 13 (a) (b) (c) Find curl of these vectors:

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SOLUTION TO EXAMPLE 13 (a) Use curl for Cartesian coordinate:

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(b) Use curl for Circular cylindrical coordinate SOLUTION TO EXAMPLE 13 (Cont’d)

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(c) Use curl for Spherical coordinate: SOLUTION TO EXAMPLE 13 (Cont’d)

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1.14 STOKE’S THEOREM The circulation of a vector field A around a closed path L is equal to the surface integral of the curl of A over the open surface S bounded by L that A and curl of A are continuous on S.

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STOKE’S THEOREM (Cont’d)

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EXAMPLE 14 By using Stoke’s Theorem, evaluate for

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EXAMPLE 14 (Cont’d)

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SOLUTION TO EXAMPLE 14 Stoke’s Theorem, where, and Evaluate right side to get left side,

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SOLUTION TO EXAMPLE 14 (Cont’d)

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EXAMPLE 15 Verify Stoke’s theorem for the vector field for given figure by evaluating: (a) over the semicircular contour. (b) over the surface of semicircular contour.

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SOLUTION TO EXAMPLE 15 (a) To find Where,

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So SOLUTION TO EXAMPLE 15 (Cont’d)

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Therefore the closed integral,

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SOLUTION TO EXAMPLE 15 (Cont’d) (b) To find

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SOLUTION TO EXAMPLE 15 (Cont’d) Therefore

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1.15 LAPLACIAN OF A SCALAR The Laplacian of a scalar field, V written as: Where, Laplacian V is:

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For Cartesian coordinate: For Circular cylindrical coordinate: LAPLACIAN OF A SCALAR (Cont’d)

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For Spherical coordinate:

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EXAMPLE 16 Find Laplacian of these scalars: (a) (b) (c) You should try this!!

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SOLUTION TO EXAMPLE 16 (a) (b) (c)

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Yea, unto God belong all things in the heavens and on earth, and enough is God to carry through all affairs Quran:4:132 END

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