lnstructions THREE submissions are allowed for each question part, except for Tr

Question

lnstructions THREE submissions are allowed for each question part, except for True/False and 2-option multipie-choicc questons. For these types or prowems, only ONE submission is allowed Exact answers are required unless the question specifically asks for a rounded answer. The types of questions that require rounded answers display the SigFig icon beside the answer box Half credit will be given on these types of problems only, for answers mat are numerically correct but do not have the correct number of significant dgits if a problem requres an exact answer, an approximated (rounded) answer will not be accepted. The MathPad or CalcPad tool that pops up will assist you on questions that require a fcorrectfy formatted mathematical expression (ie. expressions involving squareroots. Euler's number, qreek symbols, trig functions, etc.) Note that variables (letters and symbols) arc case sensitive. The message, ldquo Your answer cannot be understood or graded rdquo will appear when a syntax error (ie. missing operands, incorrect grouping operators, misspelled units, symbols with no meaning iri response) has been made. You will not lose submissions for syntax errors. Please see the topic, Answerng Math and Science Questions, under the WebAssign Studert Hetp System for detailed information and examples. Use Stokes' Theorem to evaluate integral_c F middot dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = xyi + 2zj + 7yk, C is the curve of intersection of the plane x + z = 3 and the cylinder x^2 + y^2 = 144. Submit

Explanation / Answer

Stokes theorem:
[C] P dx + Q dy + R dz =
[S] (R/y-Q/z)dydz+(P/z-R/x)dxdz+
(Q/x-P/y)dxdy, where S is a surface bounded by C.

(P,Q,R)=(xy,2z,7y).
Let S be the part of the plane x+z=3 inside the cylinder x²+y²=144.

J=[C] xy dx + 2z dy + 7y dz=
[S] (7-2)dydz+(0-0)dxdz+(0-x)dxdy=
[S1] dydz - [S2] x*dxdy=J1- J2,
where S1 and S2 are the orthogonal projections of S in y0z and x0y coordinate
planes respectively.

To find S1, you have to eliminate x, using x+z=3 and x²+y²=144 ==>
x=3-z, (3-z)²+y²=144 ==>
S1: (z-3)²+y²=12² (circle with radius=12)
J1=[S1] dydz=Area of S1
J1=144

S2: x²+y²=144
J2=[S2] x*dxdy=
[Substitute with polar coordinates: x=r*cos(t), dxdy=r*drdt, 0<t<2, 0<r<12]=
[t=0 to 2, r=0 to 12] r*cos(t)*r*drdt =
[0, 2] cos(t)*dt [0, 12] r²dr=0*[0, 12] r²dr=0

J=J1-J2=144-0
J=144

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