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Line Integrals Calculator

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Example
Created on 2024-06-20Asked by Liam King (Solvelet student)
Evaluate the line integral ∫C(2x+y) dx+(x−y) dy \int_C (2x + y) \, dx + (x - y) \, dy where C C is the curve y=x2 y = x^2 from (0,0) (0, 0) to (1,1) (1, 1) .

Solution

To evaluate the line integral ∫C(2x+y) dx+(x−y) dy \int_C (2x + y) \, dx + (x - y) \, dy where C C is the curve y=x2 y = x^2 from (0,0) (0, 0) to (1,1) (1, 1) , we parameterize the curve C C . Since y=x2 y = x^2 , let x=t x = t and y=t2 y = t^2 , where t t ranges from 0 to 1. Then dx=dt dx = dt and dy=2t dt dy = 2t \, dt . Substitute these into the integral: ∫C(2x+y) dx+(x−y) dy=∫01(2t+t2) dt+(t−t2)(2t dt) \int_C (2x + y) \, dx + (x - y) \, dy = \int_0^1 (2t + t^2) \, dt + (t - t^2)(2t \, dt) Simplify the integrand: ∫01(2t+t2) dt+∫01(2t2−2t3) dt \int_0^1 (2t + t^2) \, dt + \int_0^1 (2t^2 - 2t^3) \, dt Combine the integrals: ∫01(2t+t2+2t2−2t3) dt=∫01(2t+3t2−2t3) dt \int_0^1 (2t + t^2 + 2t^2 - 2t^3) \, dt = \int_0^1 (2t + 3t^2 - 2t^3) \, dt Evaluate the integral term by term: ∫012t dt+∫013t2 dt−∫012t3 dt \int_0^1 2t \, dt + \int_0^1 3t^2 \, dt - \int_0^1 2t^3 \, dt Compute each integral: ∫012t dt=[t2]01=1 \int_0^1 2t \, dt = \left[ t^2 \right]_0^1 = 1 ∫013t2 dt=[t3]01=1 \int_0^1 3t^2 \, dt = \left[ t^3 \right]_0^1 = 1 ∫012t3 dt=[12t4]01=12 \int_0^1 2t^3 \, dt = \left[ \frac{1}{2} t^4 \right]_0^1 = \frac{1}{2} Add the results: 1+1−12=32 1 + 1 - \frac{1}{2} = \frac{3}{2} Therefore, the value of the line integral is: 32 \frac{3}{2} Solved on Solvelet with Basic AI Model
Some of the related questions asked by Scarlett Adams on Solvelet
1. Compute the line integral ∫CF⋅dr \int_{{C}} \mathbf{F} \cdot d\mathbf{r} along the curve C C given by r(t)=(t,t2,t3) \mathbf{r}(t) = (t, t^2, t^3) , 0≤t≤1 0 \leq t \leq 1 , where F(x,y,z)=(y,x,z) \mathbf{F}(x, y, z) = (y, x, z) .2. Evaluate the line integral ∫CF⋅dr \int_{C} \mathbf{F} \cdot d\mathbf{r} along the curve C C given by r(t)=(cos⁡(t),sin⁡(t),t) \mathbf{r}(t) = (\cos(t), \sin(t), t) , 0≤t≤π 0 \leq t \leq \pi , where F(x,y,z)=(x,y,z) \mathbf{F}(x, y, z) = (x, y, z)
DefinitionLine integrals, in general, extend the idea of the integral to situations where the path of the integration matters throughout a region in the plane or space. Just as a definite integral is a number that gives us the area under a line with coordinates defined by a function, a line integral computes the sum of a scalar or a vector field along a specific curve. The line integral of a scalar field f(x,y) along a curve C, is written as ∫C​f(x,y)ds. Then with capital F being a vector field, this would be written ∫CF⋅dr. E.g. for the line segment path from (0,0) to (1,1) and field f(x,y)=x+y it would take the form ∫C(x+y)ds.
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