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Question 1 Consider the conservative vector field F = (4xy² + az²) i + (6yz + byx²)j + (cy² + 4xz) k where a, b and c are constants. i. Determine the values of a, b and c to ensure F is conservative. ii. Determine a potential function \(\phi(x, y, z)\) for F. iii. Evaluate \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where C is the curve parameterised by \(\mathbf{r}(t) = (\sin(\pi t) + 1, t \cos(\pi t), \frac{1}{2}t)\) on the interval \(0 \le t \le 2\). Question 2 Verify Stokes' theorem \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_\sigma \nabla \times \mathbf{F} \cdot d\mathbf{S}\) for the vector field \(\mathbf{F} = zx^2\mathbf{i} - 4xy\mathbf{j} + e^z\mathbf{k}\) where C is the path traced anticlockwise by a circle with radius 2 on the x-y plane centred at (2, 4, 0), \(\sigma\) is the area enclosed by C.

          Question 1
Consider the conservative vector field
F = (4xy² + az²) i + (6yz + byx²)j + (cy² + 4xz) k
where a, b and c are constants.
i. Determine the values of a, b and c to ensure F is conservative.
ii. Determine a potential function \(\phi(x, y, z)\) for F.
iii. Evaluate \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where C is the curve parameterised by
\(\mathbf{r}(t) = (\sin(\pi t) + 1, t \cos(\pi t), \frac{1}{2}t)\) on the interval \(0 \le t \le 2\).
Question 2
Verify Stokes' theorem
\(\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_\sigma \nabla \times \mathbf{F} \cdot d\mathbf{S}\)
for the vector field
\(\mathbf{F} = zx^2\mathbf{i} - 4xy\mathbf{j} + e^z\mathbf{k}\)
where C is the path traced anticlockwise by a circle with radius 2 on the x-y plane
centred at (2, 4, 0), \(\sigma\) is the area enclosed by C.

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Question 1
Consider the conservative vector field
F = (4xy² + az²) i + (6yz + byx²)j + (cy² + 4xz) k
where a, b and c are constants.
i. Determine the values of a, b and c to ensure F is conservative.
ii. Determine a potential function ϕ(x, y, z) for F.
iii. Evaluate 𝐅· d𝐫, where C is the curve parameterised by
𝐫(t) = (sin(π t) + 1, t cos(π t), (1)/(2)t) on the interval 0 ≤ t ≤ 2.
Question 2
Verify Stokes' theorem
𝐅· d𝐫 = ∬σ∇×𝐅· d𝐒
for the vector field
𝐅 = zx^2𝐢 - 4xy𝐣 + e^z𝐤
where C is the path traced anticlockwise by a circle with radius 2 on the x-y plane
centred at (2, 4, 0), σ is the area enclosed by C.

Added by Joshua G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

01/02/2023

Step 1

Step 1: To determine the values of a, b, and c to ensure F is conservative, we need to check if the curl of F is zero. Show more…

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help with either parts would be amazing! I'm a bit stuck Question 1 Consider the conservative vector field F = (4xy2+ az2) i+ (6yz+ byx2)j+ (cy2+4xz) k where a, b and c are constants. i. Determine the values of a, b and c to ensure F is conservative ii. Determine a potential function (, y, z) for F. iii. Evaluate F : dr, where C is the curve parameterised by sin(Tt)+1,t cos(Tt) on the interval 0< t<2 Question 2 Verify Stokes' theorem V xF.dS for the vector field F=zx2i-4xyj+ek where C is the path traced anticlockwise by a circle with radius 2 on the x-y plane centred at (2,4,0), is the area enclosed by C.

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00:01 Hello students, today we will discuss about this question.

00:04 In this question, we are given that we need to show that f that is equals to e -ris to x, k -z, i -carat, plus x -z minus e -race -to -x, sine -y, j -carat plus x -y -k -carat, is conservative and over its natural domain, and we need to find the potential function for it.

00:39 So here, first of all, we can write dale f that is equals to capital f.

00:47 So therefore i, dill divided by dale x plus j, del divide by del y plus k, dahl divide by dail z, that is equals to e -race to x, cos y plus y z, i carrot plus x z minus e raise to x sine y j carrot plus x y c x y plus x y plus z k so therefore d divide by dale x of f that is equals to e r to x cos y plus y z f that is equal to integration erase to x cos y plus y z d x that is equals to e, rise to x, cosy, plus x, y, z, plus g of y, z.

01:34 Now, we will integral with respect to x.

01:38 So the partial differentiation with respect to y, that is, del f, divide by del y, that is equals to x z minus e, rise to x, sine y, plus g dash of y z.

01:52 And x z minus e rise to x sine y that is equals to x z minus e rise to x x x x minus e to x x x to x x minus e rise to x x x x x of y z that is equals to 0 integral to each side with respect to y so therefore we can write dale f divided by del y that is equals to x z minus eris to x z sine y plus g dash of y z x z minus e to x sine y that is equals to x z minus e to x to x sine y plus g dash of x y z so now here g of y z that is equals to h z so put g of y z is equals to h z in equation f is equals to e, raise to x, cosy plus x, y, z plus g of y z...

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