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is 4/3 (x 1) and 2/5 (y 1) and 0 for the other faces. 1 8. r a, cos 1, V _ a(4 a 2) 3 10. Team Project. (a) Put g in (8). (b) Use (a). (c) Use (9). (d) h g is harmonic and h/ n 0 on S. Thus h const in T by (b). 2 (e) Use div (grad ) and (2). SECTION 10.9. Stokes s Theorem, page 468 Purpose. To prove, explain, and apply Stokes s theorem, relating line integrals over closed curves and surface integrals. Main Content Formula (2) (2*) Further interpretation of the curl (see also Sec. 9.9) Path independence of line integrals (leftover from Sec. 10.2) Comment on Orientation Since the choice of right-handed or left-handed coordinates is essential to the curl (Sec. 9.9), surface orientation becomes essential here (Fig. 251). Comment on Proof The proof is simple: 1. Cut (2*) into components. Take the first, (3). 2. Using N1 and N3, cast the left side of (3) into the form (7). 3. Transform the right side of (3) by Green s theorem in the plane into a double integral and show equality with the integral obtained on the left. SOLUTIONS TO PROBLEM SET 10.9, page 473 2. r [2 cos u, 2 sin u, v], curl F N Hence (curl F) N 5(cos v) 2 sin u. Integretion over u from 0 to gives 20 cos v. Integration of this over v from 0 to _ gives 20. Answer: 20. 1 2 4. curl F [ sinh z, 1, 3 sin y], N [0, 0, 1]. Hence (curl F) N 3 sin y. [0, 5 cos z, 0] 2 sin u, 0]. [0, 5 cos v, 0]. Also

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ptLineSeg[1].y = ptLineSeg[0].y; DrawLines( hWnd ); break; case IDM_FILE_UP: //set the endpoint of the line segment ptLineSeg[1].x = ptLineSeg[0].x; ptLineSeg[1].y = ptLineSeg[0].y - iSegLen; DrawLines( hWnd ); break; case IDM_FILE_DOWN: //set the endpoint of the line segment ptLineSeg[1].x = ptLineSeg[0].x; ptLineSeg[1].y = ptLineSeg[0].y + iSegLen; DrawLines( hWnd ); break; case IDM_FILE_EXIT: DestroyWindow(hWnd); break; default: return DefWindowProc(hWnd, message, wParam, lParam);

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