The line integral ∫CF⃗ ⋅dr⃗ ∫𝐶𝐹→⋅𝑑𝑟→, where F⃗ =xyi⃗ +yzj⃗ +zxk⃗ 𝐹→=𝑥𝑦𝑖→+𝑦𝑧𝑗→+𝑧𝑥𝑘→ and C𝐶 is the curve given by r⃗ (t)=ti⃗ +t2j⃗ +t3k⃗ 𝑟→(𝑡)=𝑡𝑖→+𝑡2𝑗→+𝑡3𝑘→, 0≤t≤10≤𝑡≤1 is expressed as the definite integral
Question
The line integral ∫CF⃗ ⋅dr⃗ ∫𝐶𝐹→⋅𝑑𝑟→, where F⃗ =xyi⃗ +yzj⃗ +zxk⃗ 𝐹→=𝑥𝑦𝑖→+𝑦𝑧𝑗→+𝑧𝑥𝑘→ and C𝐶 is the curve given by r⃗ (t)=ti⃗ +t2j⃗ +t3k⃗ 𝑟→(𝑡)=𝑡𝑖→+𝑡2𝑗→+𝑡3𝑘→, 0≤t≤10≤𝑡≤1 is expressed as the definite integral
Solution
To solve the line integral ∫CF⃗ ⋅dr⃗, we first need to parameterize the vector field F⃗ and the curve C.
The vector field F⃗ =xyi⃗ +yzj⃗ +zxk⃗ is given in terms of x, y, and z. We can express this in terms of t using the parameterization of the curve C, r⃗ (t)=ti⃗ +t^2j⃗ +t^3k⃗.
So, x = t, y = t^2, and z = t^3. Substituting these into F⃗ gives us F⃗(t) = tti⃗ + t^2t^3j⃗ + tt^3k⃗ = t^2i⃗ + t^5j⃗ + t^4*k⃗.
Next, we need to find dr⃗/dt, the derivative of r⃗(t) with respect to t. This gives us dr⃗/dt = i⃗ + 2tj⃗ + 3t^2k⃗.
The line integral ∫CF⃗ ⋅dr⃗ is then given by the integral from 0 to 1 of F⃗(t) ⋅ dr⃗/dt dt.
This is a dot product, so we multiply the corresponding components of F⃗(t) and dr⃗/dt and add them together:
∫ from 0 to 1 of (t^2i⃗ + t^5j⃗ + t^4k⃗) ⋅ (i⃗ + 2tj⃗ + 3t^2*k⃗) dt = ∫ from 0 to 1 of (t^2 + 2t^6 + 3t^6) dt = ∫ from 0 to 1 of (t^2 + 5t^6) dt.
This is a standard integral that can be solved using the power rule for integration, giving the final answer as [t^3/3 + t^7/7] from 0 to 1 = 1/3 + 1/7 = 10/21.
Similar Questions
<p>To evaluate the line integral ∫CF · dr, we first need to find the vector field F(x, y, z) and the vector function r(t).</p> <p>Given F(x, y, z) = (x + y^2) i + xz j + (y + z) k and r(t) = t^2 i + t^3 j - 2t k, we can find the derivative of r(t) as dr/dt = 2t i + 3t^2 j - 2 k.</p> <p>Next, we substitute r(t) into F(x, y, z) to get F(r(t)) = (t^2 + (t^3)^2) i + t^2 * (-2t) j + (t^3 - 2t) k = (t^2 + t^6) i - 2t^3 j + (t^3 - 2t) k.</p> <p>Then, we find the dot product of F(r(t)) and dr/dt:</p> <p>F(r(t)) · dr/dt = (t^2 + t^6) * 2t + (-2t^3) * 3t^2 + (t^3 - 2t) * (-2) = 2t^3 + 2t^7 - 6t^5 + 2t^3 - 4t = 2t^7 - 6t^5 + 4t^3 - 4t.</p> <p>Finally, we evaluate the line integral from t=0 to t=2:</p> <p>∫ from 0 to 2 [2t^7 - 6t^5 + 4t^3 - 4t] dt = [1/4 * t^8 - t^6 + t^4 - 2t^2] from 0 to 2 = 1/4 * (2^8) - (2^6) + (2^4) - 2*(2^2) - (0) = 64 - 64 + 16 - 8 = 8.</p> <p>So, the line integral ∫CF · dr = 8.</p> ####
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