Koja je površina ispod polarne krivulje f (theta) = theta-thetasin ((7theta) / 8) -cos ((5theta) / 3 + pi / 3) iznad [pi / 6, (3pi) / 2]?

Odgovor:

#color (crveno) ("Područje A" = 25.303335481 "" kvadratnih jedinica ") #

Obrazloženje:

Za polarne koordinate, formula za područje A:

dan # r = theta-theta * sin ((7ta)) / 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 ((7ta) / 8) + cos ^ 2 ((5ta) pi / 3) *

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

Nakon neke trigonometrijske transformacije i integracije po dijelovima slijedi

# 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 ((7theta) / 4) + theta / 2 + 3/20 x sin ((10theta) / 3 + (2pi) / 3) *

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

# 24 + / 19 * * theta cos ((19theta) / 24 + pi / 3) -576/361 * sin ((19theta) / 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 (crveno) ("Područje A" = 25.303335481 "" kvadratnih jedinica ") #

Bog blagoslovi …. Nadam se da je objašnjenje korisno.