Find-the-area-generated-when-the-curve-x-a-sin-1-cos-0-pi-rotates-about-x-axis-through-2pi-radian-Note-1-cos-2-sin-2-2-

Question Number 12577 by tawa last updated on 26/Apr/17

Find the area generated when the curve x = a (θ - \sin θ), (1 - \cos θ)

θ = 0, θ = π rotates about x - axis through 2 π radian .

Note : 1 - \cos θ = 2 \sin^{2} (\frac{θ}{2})

Answered by mrW1 last updated on 26/Apr/17

x = a (θ - \sin θ)

y = 1 - \cos θ

A = 2 π \int_{x_{1}}^{x_{2}} y d x = 2 π a \int (1 - \cos θ) (1 - \cos θ) d θ

= 2 π a \int {4 \sin}^{4} \frac{θ}{2} d θ

= 16 π a \int_{0}^{π} \sin^{4} \frac{θ}{2} d \frac{θ}{2}

= 16 π a \int_{0}^{\frac{π}{2}} \sin^{4} t d t

= 16 π a {[\frac{\sin 4 t - 8 \sin 2 t + 12 t}{32}]}_{0}^{\frac{π}{2}}

= 16 π a \frac{12}{32} \times \frac{π}{2} = 3 a π^{2}

Commented by tawa last updated on 26/Apr/17

God bless you sir .

Answered by ajfour last updated on 26/Apr/17

A r e a g e n e r a t e d = 2 π \int_{0}^{x_{1}} y d x

A = 2 π \int_{0}^{π} a {(1 - \cos θ)}^{2} d θ \dots (i)

= 2 π {\int_{0}}^{π} a {(1 + \cos θ)}^{2} d θ \dots (i i)

(i) + (i i) g i v e s :

2 A = 2 π a \int_{0}^{π} (2 + 2 \cos^{2} θ) d θ

A = π a \int_{0}^{π} (3 + \cos 2 θ) d θ

= π a {[3 θ + \frac{\sin 2 θ}{2}]}_{0}^{π}

A = 3 π^{2} a .

Commented by tawa last updated on 26/Apr/17

God bless you sir

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