Find the volume V obtained by rotating the region bounded by the curves about th

Question

Find the volume V obtained by rotating the region bounded by the curves about the given axis. y = sec(x), y = cos(x), 0 lessthanorequalto x lessthanorequalto pi/3: about y = -1 Household electricity in a particular country is supplied in the form of alternating current that varies from 175 V to -175 V with a frequency of 30 cycles per second (Hz). The voltage is thus given by the equation E(t) = 175 sin(60 pi t) where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the squareroot of the average value of [E(t)]^2 over one cycle. (a) Calculate the RMS voltage of household current in this particular country. (Round your answer to the nearest whole number of volts.) (b) Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude A needed for the voltage E(t) = A sin(60 pi t). (Round your answer to the nearest whole number of volts.)

Explanation / Answer

y=sec(x) , y=cos(x) ,0<=x<=/3 ; about y=-1

using washer method:

outer radius =sec(x) -(-1) =1+sec(x)

inner radius =cos(x) -(-1) =1+cos(x)

volume V=[0 to /3][(1+sec(x))2-(1+cos(x))2] dx

volume V=[0 to /3][2sec(x) +sec2(x)-2cos(x)-cos2(x)] dx

volume V=[0 to /3][2sec(x) +sec2(x)-2cos(x)-(1/2)-(1/2)cos(2x)] dx

volume V=[0 to /3][2ln|sec(x)+tan(x)| +tan(x)-2sin(x)-(x/2)-(1/4)sin(2x)]

volume V=[2ln|sec(/3)+tan(/3)| +tan(/3)-2sin(/3)-(/6)-(1/4)sin(2/3)] -[2ln|sec(0)+tan(0)|+tan(0)-2sin(0)-(0/2)-(1/4)sin(0)]

volume V=[2ln|2+3| +3- 3-(/6)-((3)/8)] -[2ln|1+0|+0-0-0-0]

volume V=[2ln|2+3| -(/6)-((3)/8)] -[0+0-0-0-0]

volume V=(/48)[96ln(2+3) -8- 63] 5.94958

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