Evaluate-s-F-n-dS-where-F-4xi-2y-2-j-z-2-k-and-S-is-the-surface-of-the-cylinder-bounded-by-x-2-y-2-4-z-0-and-z-3-

Question Number 100320 by bemath last updated on 26/Jun/20

Evaluate \int \int_{s} \vec{F} . \hat{n} dS where \vec{F} = 4 x \hat{i} - {2 y}^{2} \hat{j} + z^{2} \hat{k}

and S is the surface of the cylinder

bounded by x^{2} + y^{2} = 4, z = 0 and z = 3 .

Answered by smridha last updated on 26/Jun/20

\int \int_{s} \vec{F} . \hat{n d s} = \int \int_{R} \vec{F} . \hat{n} \frac{d x d z}{∣ \hat{n} . \hat{j} ∣} [t a k e a p r o j e c t i o n o n x - z p l a n e]

o u r s c a l e r p o i n t f^{n} Φ = x^{2} + y^{2} - 4

s o t h e u n i t n o r m a l o f t h i s s u r f a c e

\hat{n} = \frac{▽ Φ}{∣ ▽ Φ ∣} = \frac{x \hat{i} + y \hat{j}}{2} s o ∣ \hat{n} . \hat{j} ∣= \frac{y}{2}

a n d \vec{F} . \hat{n} = \frac{4 x^{2} - 2 y^{3}}{2} .

n o w o u r i n t e g r a l l o o k s l i k e

\int_{0}^{3} \int_{0}^{2} [\frac{4 x^{2}}{y} - 2 y^{2}] d x d z

= \int_{0}^{3} d z [\int_{0}^{2} (\frac{4 x^{2}}{\sqrt{4 - x^{2}}} + (2 x^{2} - 8)) d x]

= 3 [- 4 . \int_{0}^{2} {\sqrt{4 - x^{2}} - \frac{4}{\sqrt{4 - x^{2}}}} d x + {(\frac{2}{3} x^{3} - 8 x)}_{0}^{2}]

= 3 [- 4 {(\frac{x \sqrt{4 - x^{2}}}{2} - {2 \sin}^{- 1} (\frac{x}{2}))}_{0}^{2} + \frac{16}{3} - 16]

= 3 [4 π - \frac{32}{3}] = [12 π - 32]

Commented by bemath last updated on 26/Jun/20

great

Facebook Tweet Pin