Problem 34 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{0}^{1} \int_{y}^{2-y} \int_{0}^{2-x-y} x y d z d x d y$$

Step-by-Step Solution

Answer

The value of the given triple integral is $$\frac{1}{6}$$.

Step 1: Set up the Integral

We have a triple integral in the given order: $$\int_{0}^{1} \int_{y}^{2-y} \int_{0}^{2-x-y} x y dz dx dy$$ Let's first integrate with respect to \(z\).

Step 2: Integrate with respect to \(z\)

As \(x\) and \(y\) are constants with respect to \(z\), we have: $$\int_{0}^{2-x-y} x y d z = x y \int_{0}^{2-x-y} d z$$ Integrating with respect to \(z\) gives us: $$x y ( z \big|_{0}^{2-x-y}) = x y (2 - x - y)$$ Now we have a double integral left: $$\int_{0}^{1} \int_{y}^{2-y} x y (2 - x - y) dx dy$$

Step 3: Integrate with respect to \(x\)

Now we integrate the inner integral with respect to \(x\): $$\int_{y}^{2-y} x y (2 - x - y) dx$$ Let's first multiply the terms: $$\int_{y}^{2-y} (2 x y - x^2 y - x y^2) dx$$ Now we can integrate term by term with respect to \(x\): $$\left[2xy^2\frac{x^2}{2} - \frac{x^3y}{3} - \frac{x^2y^2}{2}\right]_{y}^{2-y}$$ Now, we will substitute the limits and simplify the expression: $$\left(4y^2 - y^3 - 2y^3\right) - \left(y^4 - \frac{y^5}{3} - y^4\right)$$

Step 4: Integrate with respect to \(y\)

We now have the final integral to solve: $$\int_{0}^{1} (3y^2 - y^3 - y^5 + \frac{y^5}{3}) dy$$ Integrating each term with respect to \(y\) gives: $$\left[ y^3 - \frac{y^4}{4} - \frac{y^6}{6} + \frac{y^6}{18} \right]_{0}^{1}$$ Substitute the limits and simplify: $$\left( 1 - \frac{1}{4} - \frac{1}{6} + \frac{1}{18} \right) - (0)$$ Simplifying further: $$\frac{9}{18} - \frac{4}{18} - \frac{3}{18} + \frac{1}{18} = \frac{3}{18}$$ The value of the triple integral is: $$\boxed{\frac{1}{6}}$$.