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# MAT 103 1.0 iafgdalaiaf.a m%fushh. iafgdalaiaf.a m%fushh f uys mDIaG wkql,fha tall wNs,usnh i|yd Ok osYdj, c odrh osf.a jdudj¾: osYdjg.uka lrk úg iqr;a.

## Presentation on theme: "MAT 103 1.0 iafgdalaiaf.a m%fushh. iafgdalaiaf.a m%fushh f uys mDIaG wkql,fha tall wNs,usnh i|yd Ok osYdj, c odrh osf.a jdudj¾: osYdjg.uka lrk úg iqr;a."— Presentation transcript:

MAT 103 1.0 iafgdalaiaf.a m%fushh

iafgdalaiaf.a m%fushh f uys mDIaG wkql,fha tall wNs,usnh i|yd Ok osYdj, c odrh osf.a jdudj¾: osYdjg.uka lrk úg iqr;a moaO;shla idok osYdjg fjs. S h kq ir, ixjD; odrhla iys; mDIaGhla hehs.ksuq. ika;;sl wjl, ix.=Kl iys; A ffoYsl lafIa;%hla i|yd fuys c hkq S mDIaGfha odrh fjs. Eg: ffoYsl lafIa;%h iy mDIaGh ie,lSfuka iafgdalaiaf.a m%fushh i;Hdmkh lrkak.

Stokes’s Theorem For a vector field A, with continuous partial derivatives Let S be a surface and, c be the simple closed curved edge of S. Here, positive direction for the surface integral is the unit vector making the right handed system when we move in anticlockwise direction along the curve.

Note:A line integral and a surface integral are connected by Stokes’s Theorem. Eg: Verify the Stokes’s Theorem with and the surface. C : Where.

Hence, Theorem is verified.

Verify the Stokes’s Theorem with and the surface. E.g.

Hence, Theorem is verified. E.g. Evaluate for The plane z = 2 Let

By Stokes theorem = o.

along the boundaries of the rectangle E.g. Evaluate Choose π

o π 1

Evaluate where c is the curve given by and begins at the point and goes at first below the Z-plane. is same as A sphere of radius centered at ( a, a, o) Further, the given plane x + y =2a passes through the center of the sphere. So it divide the sphere into two equal halves.

( 1, 0) ( -1, 0) ( 0, 1) ( 0, -1) E.g. Evaluate taken round the square C with vertices ( 1, 0),(-1, 0), ( 0, 1) and (0, -1)

Q 22 ( Page 881 ) D

Q 23 O A ( 1, 0 ) B ( 0, 1 ) By symmetry

Divergence Theorem LLet S be a closed surface and V be the volume enclosed by S. Then outward flux of continuously differential Vector Field A over S is same as the volume integral of div A. ii. e wmidrs;d m%fushh A hkq hus ixjD; mDIaGhla ;=< iy u; ika;;slj wjl,H ffoYsl flIa;%hla kus, tu mDIaGh msrsjid msg;g A ys i%djh, mDIaGfhka wdjD; jk m%foaYh msrsjid wmid A ys wkql,hg iudk fjs.

The Divergence Theorem connects a volume integral and an integral over a closed surface. Letting. A( plane surface) S o ( curve surface) Note : E.g. verify the Divergence Theorem for

For the curved surface S o Hence,

Further, gives Hence, the Divergence Theorem is verified. E.g. Evaluate S is the closed surface bounded by the cone and the plane

Solution. X Y Z

* Evaluate. where S is the surface of the cube * Suppose S is the surface of the sphere of radius a centered at O, evaluate. * For any closed surface S, evaluate. Here, Divergence Theorem can not be applied. Why? Result If A.c = o for any constant arbitrary vector c, then A = o.

1. If is a scalar field show that. Solution : Let be an arbitrary constant vector. Apply divergence Theorem to

. 2. For any closed surface s, show that Solution : Let be an arbitrary constant vector.

Since gives us Hence,.

E.g. If s is the surface of the sphere Show that Solution :

* Evaluate. where S is the surface of the cube X Y Z Here Apply divergence Theorem

D h kq Oxy ;,fha jQ c ir, ixjD; jl%fhka wdjD; jk fmfoi ku\$, D ys.=re;aj flakao%h. D Eg :

Q 23 O A ( 1, 0 ) B ( 0, 1 ) By symmetry

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