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Vector CalculusQuestion and Answers: Page 3

Question Number 128823 Answers: 0 Comments: 1

Question Number 128691 Answers: 0 Comments: 1

find∫∫_S ▽×F .n^ ds where F^→ =y^2 i^ +yj^ −xzk^ and S is the upper half of the sphere x^2 +y^2 +z^2 =a^2

$f i n d \int \int_{S} ▽ \times F . \hat{n d s}$ $w h e r e F^{\to} = y^{2} \hat{i} + y \hat{j} - x z \hat{k}$ $a n d S i s t h e u p p e r h a l f o f t h e s p h e r e$ $x^{2} + y^{2} + z^{2} = a^{2}$

Question Number 128688 Answers: 0 Comments: 0

find the flux of the vector field F=xi^ +yj^ +(√(x^2 +y^2 −1 k^ )) through outer side of hyper−boloide z=(√(x^2 +y^2 −1)) bounded by the planes z=0 and z=(√3)

$f i n d t h e f l u x o f t h e v e c t o r f i e l d$ $F = x \hat{i} + y \hat{j} + \sqrt{x^{2} + y^{2} - 1 \hat{k}}$ $t h r o u g h o u t e r s i d e o f h y p e r - b o l o i d e$ $z = \sqrt{x^{2} + y^{2} - 1}$ $b o u n d e d b y t h e p l a n e s$ $z = 0 a n d z = \sqrt{3}$

Question Number 128592 Answers: 1 Comments: 0

∫∫∫_V ▽×F dV where F=(x+2y)i^ −3zj^ +xk^ and V is the closed region in first octant by the plane 2x+2y+2z=4

$\int \int \int_{V} ▽ \times F d V w h e r e F = (x + 2 y) \hat{i} - 3 z \hat{j}$ $+ x \hat{k}$ $a n d V i s t h e c l o s e d r e g i o n i n f i r s t o c t a n t$ $b y t h e p l a n e 2 x + 2 y + 2 z = 4$

Question Number 128591 Answers: 0 Comments: 1

verify the gauss divergence theorem f=(x^2 −yz)i^ +(y^2 −zx)j^ +(z^2 −xy)k^ over the region R bounded by the parallelepiped 0≤x≤a,0≤y≤b, 0≤z≤c

$v e r i f y t h e g a u s s d i v e r g e n c e t h e o r e m$ $f = (x^{2} - y z) \hat{i} + (y^{2} - z x) \hat{j} + (z^{2} - x y) \hat{k}$ $o v e r t h e r e g i o n R b o u n d e d b y t h e$ $p a r a l l e l e p i p e d 0 ⩽ x ⩽ a, 0 ⩽ y ⩽ b,$ $0 ⩽ z ⩽ c$

Question Number 128282 Answers: 1 Comments: 0

Question Number 128234 Answers: 0 Comments: 0

Question Number 126646 Answers: 1 Comments: 1

Question Number 126572 Answers: 1 Comments: 0

Question Number 126443 Answers: 0 Comments: 0

Question Number 125672 Answers: 1 Comments: 0

N=x32y in base 10. N≡0[3] and N≡0[4]. N>8329 N has four digits. Determinate N.

$N = x 32 y i n b a s e 10 .$ $N \equiv 0 [3] a n d N \equiv 0 [4] .$ $N > 8329$ $N h a s f o u r d i g i t s .$ $D e t e r m i n a t e N .$

Question Number 124682 Answers: 0 Comments: 0

Given { ((u_0 =5)),(( u_(n+1) =3u_n −4)) :} 1. show that ∀ n∈N, u_n =2+3^u_(n+1) a. Deduct that u_n is odd. c. Show that GCD(u_n ;u_(n+1) )=1 d. Deduct GCD(6+3^(1002) ;6+3^(1003) ). GCD=greatest common divisor^

$G i v e n {\begin{cases} u_{0} = 5 \\ u_{n + 1} = 3 u_{n} - 4 \end{cases}$ $1 . s h o w t h a t \forall n \in N, u_{n} = 2 + 3^{u_{n + 1}}$ $a . D e d u c t t h a t u_{n} i s o d d .$ $c . S h o w t h a t G C D (u_{n}; u_{n + 1}) = 1$ $d . D e d u c t G C D (6 + 3^{1002}; 6 + 3^{1003}) .$ $G C D = g r e a t e s t c o m m o n {d i v i s o r}^{}$

Question Number 122358 Answers: 1 Comments: 0

Given a, b ∈[0;4] 309a+15c=226b 1) show that 2b≡0[3] and 3a≡1[5]

$G i v e n a, b \in [0; 4]$ $309 a + 15 c = 226 b$ $1) s h o w t h a t 2 b \equiv 0 [3] a n d 3 a \equiv 1 [5]$

Question Number 122189 Answers: 0 Comments: 0

Given that the forces F_1 , F_2 and F_3 have position vectors r_1 , r_2 and r_3 . Where, F_1 = (2i + 3i + k) N r_1 = (i + 2k) m F_2 = (3i + 2i + 5k) N r_2 = (2i − 4j + k) m F_3 = (−5i−5j −6k) N r_3 = (4i + 3j + 2k) m (a) Find the moment of F_1 about (2i + 3j + 4k) m. (b) Show that this system of forces reduces to a couple, G, and find the magnitude of this couple. (c) Check if F_1 and F_2 are concurrent.

$Given that the forces F_{1}, F_{2} and F_{3} have position vectors$ $r_{1}, r_{2} and r_{3} . Where,$ $F_{1} = (2 i + 3 i + k) N r_{1} = (i + 2 k) m$ $F_{2} = (3 i + 2 i + 5 k) N r_{2} = (2 i - 4 j + k) m$ $F_{3} = (- 5 i - 5 j - 6 k) N r_{3} = (4 i + 3 j + 2 k) m$ $(a) Find the moment of F_{1} about (2 i + 3 j + 4 k) m .$ $(b) Show that this system of forces reduces to a couple, G,$ $and find the magnitude of this couple .$ $(c) Check if F_{1} and F_{2} are concurrent .$

Question Number 121188 Answers: 0 Comments: 0

N is written 158b687a in base 10. with a<b. show that N≡2+a[4]

$N i s w r i t t e n 158 b 687 a i n b a s e 10 .$ $w i t h a < b .$ $s h o w t h a t N \equiv 2 + a [4]$

Question Number 119184 Answers: 0 Comments: 0

Nama : Attila Abiem Farhan Kelas : XI BKP Tugas KD 3.18 MTK 1. jawab : u^→ = (((−3)),((−4)),(( 12)) ) ∣u^→ ∣=(√((−3^2 )+(−4)^2 +12^2 )) =(√(9+16+144)) =(√(169))=13 2. jawab : a^→ = ((2),((−1)),(3) ) b^→ = ((7),((13)),(5) ) a^→ +b^→ = ((9),((12)),(8) ) ∣a^→ +b^→ ∣=(√(9^2 +12^2 +8^2 )) =(√(81+144+64)) =(√(289)) 3. jawab u^→ = ((7),(9),((−17)) ) v^→ = (((−2)),((−3)),((19)) ) u^→ −v^→ = ((9),((12)),((36)) ) ∣u^→ +v^→ ∣=(√(9^2 +12^2 +36^2 )) =(√(81+144+1296)) =(√(1521)) 4. jawab : A. 2a^→ = (((−6)),((−8)),((−24)) ) ∣2a^→ ∣=(√((−6)^2 +(−8)^2 +(−24)^2 )) =(√(36+64+576)) =(√(676)) =26 (1/2)b^→ = ((3),(4),((12)) ) ∣(1/2)b^→ ∣=(√(3^2 +4^2 +12^2 ))=(√(9+16+144))=(√(169))=13 B. 4a^→ +b^→ = (((−12)),((−16)),((−48)) ) + ((6),(8),((24)) ) = (((−6)),((−8)),((−24)) ) ∣4a^→ +b^→ ∣=(√((−6)^2 +(−8)^2 +(−24)^2 )) =(√(36+64+576)) =(√(676))=26 5. jawab : u^→ =8i+3k v^→ =5j−9k u^→ .v^→ =(8.0)i+(0.5)j+(3.−9)k =0+0−27k u^→ .v^→ =−27k

$N a m a : A t t i l a A b i e m F a r h a n$ $K e l a s : X I B K P$ $T u g a s K D 3.18 M T K$ $1 . j a w a b :$ $\vec{u} = (\begin{matrix} - 3 \\ - 4 \\ 12 \end{matrix})$ $∣ \vec{u} ∣= \sqrt{(- 3^{2}) + {(- 4)}^{2} + 12^{2}}$ $= \sqrt{9 + 16 + 144}$ $= \sqrt{169} = 13$ $2 . j a w a b :$ $\vec{a} = (\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}) \vec{b} = (\begin{matrix} 7 \\ 13 \\ 5 \end{matrix})$ $\vec{a} + \vec{b} = (\begin{matrix} 9 \\ 12 \\ 8 \end{matrix})$ $∣ \vec{a} + \vec{b} ∣= \sqrt{9^{2} + 12^{2} + 8^{2}}$ $= \sqrt{81 + 144 + 64}$ $= \sqrt{289}$ $3 . j a w a b$ $\vec{u} = (\begin{matrix} 7 \\ 9 \\ - 17 \end{matrix}) \vec{v} = (\begin{matrix} - 2 \\ - 3 \\ 19 \end{matrix})$ $\vec{u} - \vec{v} = (\begin{matrix} 9 \\ 12 \\ 36 \end{matrix})$ $∣ \vec{u} + \vec{v} ∣= \sqrt{9^{2} + 12^{2} + 36^{2}}$ $= \sqrt{81 + 144 + 1296}$ $= \sqrt{1521}$ $4 . j a w a b :$ $A . 2 \vec{a} = (\begin{matrix} - 6 \\ - 8 \\ - 24 \end{matrix})$ $∣ 2 \vec{a} ∣= \sqrt{{(- 6)}^{2} + {(- 8)}^{2} + {(- 24)}^{2}}$ $= \sqrt{36 + 64 + 576}$ $= \sqrt{676}$ $= 26$ $\frac{1}{2} \vec{b} = (\begin{matrix} 3 \\ 4 \\ 12 \end{matrix})$ $∣ \frac{1}{2} \vec{b} ∣= \sqrt{3^{2} + 4^{2} + 12^{2}} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ $B . 4 \vec{a} + \vec{b} = (\begin{matrix} - 12 \\ - 16 \\ - 48 \end{matrix}) + (\begin{matrix} 6 \\ 8 \\ 24 \end{matrix}) = (\begin{matrix} - 6 \\ - 8 \\ - 24 \end{matrix})$ $∣ 4 \vec{a} + \vec{b} ∣= \sqrt{{(- 6)}^{2} + {(- 8)}^{2} + {(- 24)}^{2}}$ $= \sqrt{36 + 64 + 576}$ $= \sqrt{676} = 26$ $5 . j a w a b :$ $\vec{u} = 8 i + 3 k$ $\vec{v} = 5 j - 9 k$ $\vec{u} . \vec{v} = (8.0) i + (0.5) j + (3 . - 9) k$ $= 0 + 0 - 27 k$ $\vec{u} . \vec{v} = - 27 k$