Problem 32 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{2-x} 4 y z d z d y d x$$

Step-by-Step Solution

Answer

In this problem, we needed to evaluate a triple integral of the function \(f(x, y, z) = 4yz\) over the region defined by the given limits of integration for \(x\), \(y\), and \(z\). By integrating the function first with respect to \(z\), then with respect to \(y\), and finally with respect to \(x\), we found that the value of the triple integral is \(\frac{23}{15}\).

Step 1: Integrate with respect to z

First, we will integrate the function \(f(x, y, z) = 4yz\) with respect to \(z\) over the interval \([0, 2-x]\). To do this, we'll calculate the antiderivative of \(4yz\) with respect to \(z\), and then evaluate it at the limits of integration: $$ \int_{0}^{2-x} 4yz \, dz = 2yz^2 \Big|_0^{2-x} $$ Now, let's plug in the bounds: $$ 2y(2-x)^2 - 2y(0)^2 = 2y(4 - 4x + x^2) $$

Step 2: Integrate with respect to y

Now, we have a double integral with respect to \(y\) and \(x\). Let's find the antiderivative of our current function \(2y(4 - 4x + x^2)\) with respect to \(y\) as follows: $$ \int_{0}^{\sqrt{1-x^2}} 2y(4 - 4x + x^2) \, dy = (4 - 4x + x^2)y^2 \Big|_0^{\sqrt{1-x^2}} $$ Next, plug in the bounds: $$ (4 - 4x + x^2)(1-x^2) - (4 - 4x + x^2)(0)^2 = (4 - 4x + x^2)(1-x^2) $$

Step 3: Integrate with respect to x

Finally, we will integrate our remaining function \((4 - 4x + x^2)(1-x^2)\) with respect to \(x\) over the interval \([0, 1]\): $$ \int_{0}^{1} (4 - 4x + x^2)(1-x^2) \, dx $$ To make it easier, first expand the expression: $$ \int_{0}^{1} (4-4x^3-4x+x^2+x^4) \, dx $$ Now integrate term by term: $$ \left[4x - x^4 + \frac{-4x^2}{2} + \frac{x^3}{3} + \frac{x^5 }{5}\right]_0^1 = 4 - 1 + (-2) +( \frac{1}{3}) +(\frac{1}{5}) $$ Finally, simplify the expression and find the value of the integral: $$ 4 - 1 - 2 + \frac{1}{3} + \frac{1}{5} = 1 + \frac{1}{3} + \frac{1}{5} = \frac{15}{15} + \frac{5}{15} + \frac{3}{15} = \frac{23}{15} $$ Thus, the value of the triple integral is: $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{2-x} 4 y z \, dz \, dy \, dx = \frac{23}{15} $$