Evaluate the line integral  CF · dr, where C is given by the vector function r(t).F(x, y, z) = (x + y2) i + xz j + (y + z) k,r(t) = t2 i + t3 j − 2t k,  0 ≤ t ≤ 2

<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> ####

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

Integrate f(x, y, z) = (x 2 + z 2 )0.5 over the circle, r(t) = (acost)j + (asint)k, 0 ≤ t ≤ 2π

Evaluate the following line integral over the curve given by x = t2 , y = t3 , where 1 ≤t ≤ 2

The field F = xy i + y j – yz k is the velocity field of a flow inspace. Find the flow from (0, 0, 0) to (1, 1, 1) along the curve of intersection of thecylinder y = x2 and the plane z = x shown in Figure 1. (Hint: Use t = x as theparameter.)