Question

Given that F = 2xy³i + 3x²y²j is conservative with potential function f = x²y³ + C evaluate \int_{(2,1)}^{(1,3)} 2xy³dx + 3x²y²dy. (a) 46 (b) 23 (c) 19 (d) 5 (e) 1 Problem 2 Find the divF if F = (x/y)i - (y/x)j. Problem 3 Use Green's Theorem to complete the integral statement appearing beneath to express the counterclockwise circulation of F = 3yi + 2xj around C as a double integral. Simplify the integrand. $\oint_c 3ydx + 2xdy = \iint_R ( )dA$

          Given that F = 2xy³i + 3x²y²j is conservative with potential function f = x²y³ + C evaluate \int_{(2,1)}^{(1,3)} 2xy³dx + 3x²y²dy.
(a) 46
(b) 23
(c) 19
(d) 5
(e) 1
Problem 2
Find the divF if F = (x/y)i - (y/x)j.
Problem 3
Use Green's Theorem to complete the integral statement appearing beneath to express the counterclockwise
circulation of F = 3yi + 2xj around C as a double integral. Simplify the integrand.
$\oint_c 3ydx + 2xdy = \iint_R ( )dA$
        
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Given that F = 2xy³i + 3x²y²j is conservative with potential function f = x²y³ + C evaluate ∫(2,1)^(1,3) 2xy³dx + 3x²y²dy.
(a) 46
(b) 23
(c) 19
(d) 5
(e) 1
Problem 2
Find the divF if F = (x/y)i - (y/x)j.
Problem 3
Use Green's Theorem to complete the integral statement appearing beneath to express the counterclockwise
circulation of F = 3yi + 2xj around C as a double integral. Simplify the integrand.
3ydx + 2xdy =  ( )dA

Added by Xavier H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Given that F=2xyi+3yj is conservative with potential function f=xy+C, evaluate ∫(0.32xydx+3xydy) from (a46, b23) to (c19, a3) and (e1). Problem 2: Find the divergence of F if F=x/yi-y/xj. Problem 3: Use Green's Theorem to complete the integral statement appearing beneath to express the counterclockwise circulation of F=3yi+2xj around C as a double integral. Simplify the integrand. ∮(3ydx+2xdy) = ∬Sf dA.
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Transcript

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00:01 Hi, in this equation we are given with f that is equals to x cube i cap plus y cube j cap plus z cube k cap and the surface is given as x square plus y square plus z square is equal to 4.
00:19 First we can find the divergence of f that will be equals to curly by curly x of x cube plus curly by curly y of y cube plus curly by curly z of z cube and solving this we get this to be equals to 3 times x square plus 3 times y square plus 3 times z square.
00:43 We are taking 3 as common we were left with x square plus y square plus z square.
00:49 Now as we know by divergence theorem as we get double integral over s f dot n ds is equals to triple integral over v divergence of f dv and substituting the known values here we get this to be equals to triple integral over v here we get 3 times x square plus y square plus z square times here we get dx dy dz and here since v is the volume of the sphere we transfer the triple integral into the spherical polar coordinates as we can write x square plus y square plus z square is equals to r square and dx dy dz is equals to dv and here we can replace our dv as equals to r square sine of theta dr d theta d phi...
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