Math Problem Statement

The first integral is from 0 to 1 the sencond is from 0 to 5-5x the third integral is from 0 to 5-5x-y

Evaluate the triple integral \( \int_0^1 \int_0^{5-5x} \int_0^{5-5x-y} dz \, dy \, dx \).

Solution

Thank you for uploading the image! The integral evaluates as follows:

We are given the triple integral: 01055x055xydzdydx\int_0^1 \int_0^{5-5x} \int_0^{5-5x-y} dz \, dy \, dx

Step-by-step Solution:

  1. Evaluate the innermost integral: 055xydz=[z]055xy=(55xy)0=55xy\int_0^{5-5x-y} dz = [z]_0^{5-5x-y} = (5 - 5x - y) - 0 = 5 - 5x - y

    After solving this, the integral becomes: 01055x(55xy)dydx\int_0^1 \int_0^{5-5x} (5 - 5x - y) \, dy \, dx

  2. Evaluate the second integral (in terms of yy): Expand the expression 55xy5 - 5x - y: 055x(55xy)dy=055x(55x)dy055xydy\int_0^{5-5x} (5 - 5x - y) \, dy = \int_0^{5-5x} (5 - 5x) \, dy - \int_0^{5-5x} y \, dy

    Evaluate each term:

    • For 055x(55x)dy\int_0^{5-5x} (5 - 5x) \, dy: 055x(55x)dy=(55x)y055x=(55x)(55x)=(55x)2\int_0^{5-5x} (5 - 5x) \, dy = (5 - 5x) \cdot y \Big|_0^{5-5x} = (5 - 5x)(5 - 5x) = (5 - 5x)^2

    • For 055xydy\int_0^{5-5x} y \, dy: 055xydy=y22055x=(55x)22\int_0^{5-5x} y \, dy = \frac{y^2}{2} \Big|_0^{5-5x} = \frac{(5 - 5x)^2}{2}

    Combine the results: 055x(55xy)dy=(55x)2(55x)22=(55x)22\int_0^{5-5x} (5 - 5x - y) \, dy = (5 - 5x)^2 - \frac{(5 - 5x)^2}{2} = \frac{(5 - 5x)^2}{2}

    So the integral becomes: 01(55x)22dx\int_0^1 \frac{(5 - 5x)^2}{2} \, dx

  3. Evaluate the outermost integral (in terms of xx): Expand (55x)2(5 - 5x)^2: (55x)2=2550x+25x2(5 - 5x)^2 = 25 - 50x + 25x^2

    Substitute into the integral: 01(55x)22dx=1201(2550x+25x2)dx\int_0^1 \frac{(5 - 5x)^2}{2} \, dx = \frac{1}{2} \int_0^1 (25 - 50x + 25x^2) \, dx

    Evaluate each term:

    • For 0125dx\int_0^1 25 \, dx: 0125dx=25x01=25(1)25(0)=25\int_0^1 25 \, dx = 25x \Big|_0^1 = 25(1) - 25(0) = 25
    • For 0150xdx\int_0^1 -50x \, dx: 0150xdx=50x2201=50122=25\int_0^1 -50x \, dx = -50 \cdot \frac{x^2}{2} \Big|_0^1 = -50 \cdot \frac{1^2}{2} = -25
    • For 0125x2dx\int_0^1 25x^2 \, dx: 0125x2dx=25x3301=25133=253\int_0^1 25x^2 \, dx = 25 \cdot \frac{x^3}{3} \Big|_0^1 = 25 \cdot \frac{1^3}{3} = \frac{25}{3}

    Combine the results: 01(2550x+25x2)dx=2525+253=253\int_0^1 (25 - 50x + 25x^2) \, dx = 25 - 25 + \frac{25}{3} = \frac{25}{3}

    Multiply by 12\frac{1}{2}: 01(55x)22dx=12253=256\int_0^1 \frac{(5 - 5x)^2}{2} \, dx = \frac{1}{2} \cdot \frac{25}{3} = \frac{25}{6}

Final Answer:

01055x055xydzdydx=256\int_0^1 \int_0^{5-5x} \int_0^{5-5x-y} dz \, dy \, dx = \frac{25}{6}

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Follow-up Questions:

  1. How do we interpret triple integrals geometrically?
  2. What does the region of integration look like in this example?
  3. Can the order of integration be changed for this problem?
  4. How would we evaluate this integral using numerical methods?
  5. What are common applications of triple integrals in physics or engineering?

Tip:

Always check the bounds of integration to ensure they describe a valid geometric region before evaluating a multiple integral!

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Math Problem Analysis

Mathematical Concepts

Triple Integrals
Integration Techniques
Iterated Integration

Formulas

Integral of a constant: \( \int_a^b c \, dx = c(b-a) \)
Integral of a polynomial: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
Sum Rule for Integrals: \( \int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx \)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Calculus