Question

Fundamental Theorem of Line Integrals Consider the line integral: ∮_C▒〈6xy+6x, 3x^2-3 cos⁡(y) 〉 ∙dr ⃗ where C is the line segment from the origin to (4, 5). Evaluate this integral by rewriting the integral as a standard single variable integral of the parameter t. Without finding a potential function show that the integrand is a gradient field. Find a potential function for the vector field. Use the Fundamental Theorem of Line Integrals to evaluate this integral.

Answer 1

C is the line segment from the origin to (4, 5)

r(t)=〈4t,5t〉 ,0≤t≤1
=>r'(t)=〈4,5〉

F=〈6xy+6x, 3x²-3cos⁡(y)〉
=>F(r(t))=〈6(4t)(5t)+6(4t), 3(4t)²-3cos⁡(5t)〉
=>F(r(t))=〈120t²+24t, 48t²-3cos⁡(5t)〉

∮_C▒〈6xy+6x, 3x^2-3 cos⁡(y) 〉 ∙dr ⃗
=₀∫¹F(r(t)) .r'(t) dt
=₀∫¹〈120t²+24t, 48t²-3cos⁡(5t)〉.〈4,5〉 dt
=₀∫¹[(120t²+24t)(4)+(48t²-3cos⁡(5t))(5)] dt
=₀∫¹[480t²+96t+240t²-15cos⁡(5t)] dt
=₀∫¹[720t²+96t-15cos⁡(5t)] dt
=[240t³+48t²-3sin⁡(5t)]  ₀|¹
=[240(1)³+48(1)²-3sin⁡(5*1)]-[240(0)³+48(0)²-3sin⁡(5*0)]
=[240+48-3sin⁡(5)]-[0+0-3sin⁡(0)]
=288-3sin⁡(5)

-----------------------------------------------------------------------------------------

f_x=6xy+6x, f_y=3x²-3cos⁡(y)

f=∫f_x dx
=>f=∫(6xy+6x) ​​​​​dx
=>f=3x²y+3x²+g(y) ​​​
=>f_y=3x²+0+g'(y) ​
=>f_y=3x²+g'(y) ​

3x²+g'(y) ​=3x²-3cos⁡(y)
=>g'(y) ​=-3cos⁡(y)
=>g(y) ​=∫-3cos⁡(y) dy
=>g(y) ​=-3sin⁡(y) +c

potential function f(x,y)=3x²y+3x²-3sin⁡(y) +c

Using the Fundamental Theorem of Line Integrals

∮_C▒〈6xy+6x, 3x^2-3 cos⁡(y) 〉 ∙dr ⃗

=f(4,5)-f(0,0)
=[3(4)²(5)+3(4)²-3sin⁡(5) +c]-[3(0)²(0)+3(0)²-3sin⁡(0) +c]
=[240+48-3sin⁡(5) +c]-[0+0-0 +c]
=288-3sin⁡(5)