Math Problem Statement

Обчислити ∫(x + y + z)dl, де L - відрізок прямої між A(0,0,0) та B(1,1,1).

Solution

The problem asks to compute the line integral:

L(x+y+z)dl\int_L (x + y + z) \, dl

where LL is the line segment connecting points A(0,0,0)A(0,0,0) and B(1,1,1)B(1,1,1).

Solution:

The line segment LL between A(0,0,0)A(0,0,0) and B(1,1,1)B(1,1,1) can be parametrized as:

r(t)=(t,t,t),t[0,1]\mathbf{r}(t) = (t, t, t), \quad t \in [0, 1]

This is because x=tx = t, y=ty = t, and z=tz = t for t[0,1]t \in [0, 1].

Now, we can calculate the differential line element dldl. The magnitude of the derivative of r(t)\mathbf{r}(t) is:

dl=r(t)dt=12+12+12dt=3dtdl = \|\mathbf{r}'(t)\| \, dt = \sqrt{1^2 + 1^2 + 1^2} \, dt = \sqrt{3} \, dt

Now, substitute the values of x=tx = t, y=ty = t, and z=tz = t into the integrand:

x+y+z=t+t+t=3tx + y + z = t + t + t = 3t

Thus, the integral becomes:

01(3t)3dt=3301tdt\int_0^1 (3t) \cdot \sqrt{3} \, dt = 3\sqrt{3} \int_0^1 t \, dt

The integral of tt is:

01tdt=t2201=12\int_0^1 t \, dt = \frac{t^2}{2} \Big|_0^1 = \frac{1}{2}

So, the result is:

3312=3323\sqrt{3} \cdot \frac{1}{2} = \frac{3\sqrt{3}}{2}

Final Answer:

332\frac{3\sqrt{3}}{2}

Let me know if you need further details or have any questions!


Here are 5 related questions to expand on this topic:

  1. How would the integral change if the line segment was not linear but a curve?
  2. Can you parametrize a different line integral between two arbitrary points?
  3. What is the physical interpretation of a line integral in this context?
  4. How do you compute a surface integral over a plane or curved surface?
  5. What is the difference between a line integral and a regular definite integral?

Tip: For line integrals, always ensure you correctly parametrize the path and compute the differential length dldl accordingly.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Line Integrals
Parametric Equations

Formulas

Line segment parametrization: r(t) = (t, t, t), where t ∈ [0, 1]
Line element: dl = √(1^2 + 1^2 + 1^2) dt = √3 dt
Integral calculation: ∫(x + y + z) dl

Theorems

Parametric representation of line integrals
Basic integration of polynomials

Suitable Grade Level

University - Calculus III