Vector Calculus line integral help?
To find c(t), you will need to parametrize the curve. Since y = x^3 for 0 <= x <= 1, we can use x = t, y = t^3, 0 <= t <= 1. Thus we can use c(t) =
How do I calculate value of curve integral?
The infinitesimal element of distance is [math]ds=\sqrt{dx^{2}+dy^{2}}[/math]. Here, [math]x=t\cos t\implies dx=dt\cos t-t\sin t dt=(\cos t-t\sin t)dt[/math]. [math]y=t\sin t\implies dy=dt\sin t+t\cos t dt=(\sin t+t\cos t)dt[/math]. Then [math]ds=\sqrt{dx^{2}+dy^{2}}=\sqrt{t^{2}+1}dt. \sqrt{x^{2}+y^{2}+1}=\sqrt{t^{2}+1}[/math]. So, [math]\int_{C}f ds=\int_{0}^{4\pi}\frac{4t^{2}\cos^{2}t}{\sqrt{t^{2}+1}}dt=\frac{64}{3}\pi^{3}+4\pi[/math].Refer to https://www.wolframalpha.com for the curve.
Compute the value of the line integral of the vector field v over the given oriented, closed curve C.?
First the hard way, then the easy way using Green's theorem: Express each of the three segments in parametric form: ∫ C = ∫ C₁ + ∫ C₂ + ∫ C₃ C₁ = (1 - 2t) i .......... (1,0) to (-1,0) t ∈ [0,1] C₂ = (t - 1)i + tj ....... (-1,0) to (0,1) t ∈ [0,1] C₃ = ti + (1 - t)j ....... (0,1) to (1,0) t ∈ [0,1] Compute each of the three segments (integrate each from 0 to 1): ∫ C₁ = ∫ v(r)·r' dr = ∫ <0,3>·<-2,0> dt = 0 ∫ C₂ = ∫ v(r)·r' dr = ∫ <(t-1)² t + t³/3,(t-1)t²>·<1,1> dt = ∫ t³/3 + (t-1)t² + (t-1)² t + 3 dt = 37/12 ∫ C₃ = ∫ v(r)·r' dr = ∫
In need of help with a Line Integral! Calculus 3?
C is parametrized by x = 4t, y = 2t^2, z = 2t^3, 0 ≤ t ≤ 1. Since y'(t) = 4t, we have ∫_C xye^(yz) dy = ∫[0,1] x(t) y(t) e^(y(t) * z(t)) y'(t) dt = ∫[0,1] (4t)(2t^2)e^(2t^2 * 2t^3) * 4t dt = ∫[0,1] 32t^4 e^(4t^5) dt = ∫[0,4] 32e^u * 1/20 du, using the u-sub u = 4t^5, du = 20t^4 dt = (8/5)e^u (from u = 0 to 4) = (8/5)(e^4 - 1).
Line Integrals (Calculus III)?
There are a lot of ways to do this... but the most straight forward way to approach line integrals is to find a parametric equation for your path and substitute. the the straight-line path from (0,0 to (4,4).. x = t y = t dx = dt dy = dt t goes from 0 to 4 now plug in t for x and y and dt for dx and dy ∫ 0 dt = 0 now for the path from 4,4 to 8,0 x = 4+t y = 4-t dx = dt dy = -dt t goes from 0 to 4 ∫ -8 dt -8 t -32 2) this is actually a slightly different kind of integral... there are two kinds of multivariate functions. Those were f(x,y,z) --> a scalar and those where the function maps to a vector. In the first question f(xy) = (-4, 4) When we do our path integral then we integrate ∫ f ⋅ dr Now we have a scalar output for so we will integrate ∫ f ||dr|| x = cos t y = sin t dx = -sin t dt dy = cos t dt || dr || = (dx^2 + dy^2)^1/2 which in this case = dt ∫ 3cos t - 2 sin t dt 3 sin t + 2 cos t | [0,pi] 4
Evaluate the line integral, where C is the given curve. (Round your answer to one decimal place.) ∫C xy2ds, C?
Since the time is running out, I don't have time to put up all the computation, so... Refer to this link to learn how to evaluate the line integral: http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsIntro.aspx [See examples for practice problems] in.t(x = C) (upside down triangle)F * dr... Good luck!
How do I calculate the curve integral if two curves are given?
The equation of a line joining the points [math](x_{1},y_{1},z_{1})[/math] and [math](x_{2},y_{2},z_{2}) [/math]in parametric form is given by [math]\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}=t[/math], where [math]t[/math] is a parameter such that [math]0\le t\le 1[/math]. [math]C_{1}[/math] is such a line that joins [math](x_{1},y_{1},z_{1})=(0,0,0)[/math] to [math](x_{2},y_{2},z{2})=(4,2,4)[/math]. So, the equation of [math]C_{1}[/math] is [math]\frac{x}{4}=\frac{y}{2}=\frac{z}{4}=t[/math]. So, we have [math]x=4t[/math], [math]y=2t[/math], [math]z=4t[/math]. We now express the line integral [math]\int_{C_{1}}u dx+v dy+w dz=\int_{C_{1}}(-y^{4}+4yz^{2})dx+(x^{2}z^{2}+yz)dy+(2x^{2}yz)dz[/math] as a definite integral in which [math]t[/math] is the variable of integration. So the line integral along [math]C_{1}[/math] becomes [math]\int_{t=0}^{1}(-y^{4}+4yz^{2})\frac{dx}{dt}dt+(x^{2}z^{2}+yz)\frac{dy}{dt}dt+(2x^{2}yz)\frac{dz}{dt}dt[/math], where the variables [math]x[/math], [math]y[/math],[math] z[/math], are expressed as functions of [math]t[/math], and [math]\frac{dx}{dt}=4[/math], [math]\frac{dy}{dt}=2[/math], and [math]\frac{dz}{dt}=4[/math]. As a result, the line integral is [math]\int_{t=0}^{1}[-(2t)^{4}+4(2t)(4t)^{2}]4dt+[(2t)^{2}(2t)^{2}+(2t)(4t)]2dt+[2(4t)^{2}(2t)(4t)]4dt, [/math]which can then be computed.For the curve [math]C_{2}[/math], [math]x=at^{2}[/math], [math]y=bt[/math], [math]z=ct^{3}[/math], [math]0\le t\le 1[/math]. At [math]t=0[/math], [math](x,y,z)=(0,0,0)[/math], and at [math]t=1[/math], [math](x,y,z)=(4,2,4)\implies a=4[/math], [math]b=2[/math], [math]c=4[/math]. Now, [math]\frac{dx}{dt}=2at[/math], [math]\frac{dy}{dt}=b[/math], [math]\frac{dz}{dt}=3ct^{2}[/math]. Putting the values of [math]a[/math],[math]b[/math], and [math]c[/math], [math]\frac{dx}{dt}=8t[/math], [math]\frac{dy}{dt}=2[/math], [math]\frac{dz}{dt}=12t^{2}[/math]. The line integral [math]\int_{C_{2}}udx+vdy+wdz[/math] becomes [math]\int_{t=0}^{1}[-(2t)^{4}+4(2t)(4t^{3})^{2}]8tdt+[(4t^{2})^{2}(4t^{3})^{2}+(2t)(4t^{3})]2dt+[2(4t^{2})^{2}(2t)(4t^{3})]12t^{2}dt, [/math]which can then be evaluated.
How do you compute a line integral of a vector field along a parametric curve?
Suppose that:[math]f(x,y,z)[/math] is a function that provides the vector in the vector field at some 3D coordinate [math](x,y,z)[/math].[math]x(t)[/math], [math]y(t)[/math], and [math]z(t)[/math] provide the values of [math]x[/math], [math]y[/math], and [math]z[/math] given some parameter [math]t[/math].We want to evaluate [math]\int_a^b f(x,y,z)\ dt[/math] for some values [math]a[/math] and [math]b[/math].First rewrite [math]f(x,y,z)[/math] as [math]f(t)[/math] using the functions for [math]x[/math], [math]y[/math], and [math]z[/math] in (2). Then evaluate the integral [math]\int_a^b f(t)\ dt[/math].
Evaluate the following scalar line integral where c is the given curve.
I=Integral(y/x)ds with c:x=t^4,y=t^3, 1/2 ∫c (y/x) ds = ∫(t = 1/2 to 1) (y/x) √((dx/dt)^2 + (dy/dt)^2) dt = ∫(t = 1/2 to 1) (t^3/t^4) √((4t^3)^2 + (3t^2)^2) dt = ∫(t = 1/2 to 1) (1/t) √(16t^6 + 9t^4) dt = ∫(t = 1/2 to 1) (1/t) √(t^4 (16t^2 + 9)) dt = ∫(t = 1/2 to 1) (1/t) t^2√(16t^2 + 9) dt = ∫(t = 1/2 to 1) t (16t^2 + 9)^(1/2) dt = (2/3)(1/32)(16t^2 + 9)^(3/2) {for t = 1/2 to 1} = (1/48) (125 - 13^(3/2)). I hope this helps!
Line Integral on the curve C consisting of the x axis, arc of a circle, below y=x?
Evaluate the integral over the closed curve F * dr when F = (3 x^2)i+ (2 x^2 - 3 x y^2)j on the curve C consisting of the x-axis from x=0 to x=2, the arc of the circle x^2 + y^2 = 4 up to the line y=x, and the line y=x down to the origin.