A-closed-surface-is-defined-in-spherical-coordinates-by-3-lt-r-lt-5-0-1pi-lt-lt-0-3pi-1-2pi-lt-lt-1-6pi-Find-the-total-surface-area-

Question Number 82591 by Learner-123 last updated on 22/Feb/20

A c l o s e d s u r f a c e i s d e f i n e d i n s p h e r i c a l

c o o r d i n a t e s b y 3 < r < 5, 0.1 π < θ < 0.3 π,

1.2 π < ϕ < 1.6 π . F i n d t h e t o t a l s u r f a c e

a r e a .

Commented by Learner-123 last updated on 22/Feb/20

A n s : 36.79 s q u a r e u n i t s .

Commented by Learner-123 last updated on 23/Feb/20

Commented by Learner-123 last updated on 23/Feb/20

S i r, t h i s i s t h e e x p l a i n a t i o n g i v e n .

C a n y o u t e l l h o w t h e s e 4 (a c t u a l l y 5) t e r m s

c o m e . .

Commented by mr W last updated on 23/Feb/20

Commented by mr W last updated on 23/Feb/20

i n o u r c a s e

ϕ = 0.1 π . . 0.3 π

θ = 1.2 π . . 1.6 π

r = 3 . . 5

w e h a v e t o t a l l y 6 f a c e s :

f a c e 1 : r = 3, ϕ = 0.1 π \dots 0.3 π, θ = 1.2 π . . 1.6 π

A_{1} = \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} r \sin θ d ϕ r d θ = 3^{2} \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} \sin θ d ϕ d θ

f a c e 2 : r = 5, ϕ = 0.1 π \dots 0.3 π, θ = 1.2 π . . 1.6 π

A_{2} = s i m i l a r l y = 5^{2} \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} \sin θ d ϕ d θ

f a c e 3 : ϕ = 0.1 π, θ = 1.2 π . . 1.6 π, r = 3 . . 5

A_{3} = \frac{1}{2} (5^{2} - 3^{2}) (1.6 π - 1.2 π)

f a c e 4 : ϕ = 0.3 π, θ = 1.2 π . . 1.6 π, r = 3 . . 5

A_{4} = \frac{1}{2} (5^{2} - 3^{2}) (1.6 π - 1.2 π)

f a c e 5 : ϕ = 0.1 π . . 0.3 π, θ = 1.2 π, r = 3 . . 5

A_{5} = \int_{0.1 π}^{0.3 π} \int_{3}^{5} \sin 1.2 π d ϕ r d r

f a c e 6 : ϕ = 0.1 π . . 0.3 π, θ = 1.6 π, r = 3 . . 5

A_{6} = \int_{0.1 π}^{0.3 π} \int_{3}^{5} \sin 1.6 π d ϕ r d r

A = Σ A

= (3^{2} + 5^{2}) \int_{1.2 π}^{1.6 π} \int_{0.1 π}^{0.3 π} \sin θ d ϕ d θ

+ 2 \times \frac{1}{2} (5^{2} - 3^{2}) (1.6 π - 1.2 π)

+ (\sin 1.2 π + \sin 1.6 π) (0.3 π - 0.1 π) \int_{3}^{5} r d ϕ d r

Commented by Learner-123 last updated on 24/Feb/20

T h a n k Y o u S i r!

C a n y o u p l e a s e s o l v e m y o t h e r 3 d o u b t s ?

Commented by mr W last updated on 24/Feb/20

i {d o n}^{'} t k n o w w h a t a r e y o u r d o u b t s . y o u

s h o u l d n o t a s s u m e t h a t i k n o w a l l p o s t s .

Commented by Learner-123 last updated on 24/Feb/20

Q u e s t i o n n o : 82627, 82631, 82647

Commented by Learner-123 last updated on 24/Feb/20

I t h i n k t h e r e i s s o m e t e c h n i c a l p r o b l e m

i n t h i s a p p, w h e n e v e r i p r e s s “ s o r t b y

r e c e n t {a c t i v i t y}^{″} i t g e t s m e t o s a m e

i n t e g r a l q u e s t i o n e v e r y t i m e \dots (75986)

Commented by mr W last updated on 24/Feb/20

i {c a n}^{'} t h e l p y o u .

Commented by mr W last updated on 26/Feb/20

b o t h

Commented by Learner-123 last updated on 26/Feb/20

A r e y o u r e f e r r i n g t o m y d o u b t o r

t h i s t e c h n i c a l g l i t c h ?