Kakšna je površina pod polarno krivuljo f (theta) = theta-thetasin ((7theta) / 8) -cos ((5theta) / 3 + pi / 3) nad [pi / 6, (3pi) / 2]?

Odgovor:

#color (rdeča) ("območje A" = 25.303335481 "" "kvadratnih enot") #

Pojasnilo:

Za koordinate Polar je formula za območje A:

Glede na # r = theta-theta * sin ((7theta) / 8) -cos ((5theta) / 3 + pi / 3) #

# A = 1/2 int_alpha ^ beta r ^ 2 * d theta #

# A = 1/2 int_ (pi / 6) ^ ((3pi) / 2) (theta-theta * sin ((7theta) / 8) -cos ((5theta) / 3 + pi / 3)) ^ 2 d theta #

# A = 1/2 int_ (pi / 6) ^ ((3pi) / 2) theta ^ 2 + theta ^ 2 * sin ^ 2 ((7theta) / 8) + cos ^ 2 ((5theta) / 3 + pi / 3) #

# -2 * theta ^ 2 * sin ((7theta) / 8) + 2 * theta * cos ((5theta) / 3 + pi / 3) * sin ((7theta) / 8) ## -2 * theta * cos ((5tha) / 3 + pi / 3) d theta #

Po nekaj trigonometričnih transformacijah in integraciji po delih sledi

# A = 1/2 theta ^ 3/3 + theta ^ 3 / 6-2 / 7 * theta ^ 2 * sin ((7theta) / 4) -16 / 49 * theta * cos ((7theta) / 4) + 64/343 * sin ((7ta) / 4) + theta / 2 + 3/20 * sin ((10tea) / 3 + (2pi) / 3) #

# + 16/7 * theta ^ 2 * cos ((7tha) / 8) -256 / 49 * theta * sin ((7theta) / 8) -2048 / 343 * cos ((7theta) / 8) -24/61 * theta * cos ((61theta) / 24 + pi / 3) + 576/3721 * sin ((61taj) / 24 + pi / 3) #

# + 24/19 * theta * cos ((19tea) / 24 + pi / 3) -576 / 361 * sin ((19tea) / 24 + pi / 3) ## -6 / 5 * theta * sin ((5theta) / 3 + pi / 3) -18 / 25 * cos ((5theta) / 3 + pi / 3) _ (pi / 6) ^ ((3pi) / 2) #

# A = 1/2 * 43.22026786 - (- 7.386403099) #

# A = 1/2 * (50.60667096) #

#color (rdeča) ("območje A" = 25.303335481 "" "kvadratnih enot") #

Bog blagoslovi …. Upam, da je razlaga koristna.