Problem 41 - Calculus Early Transcendentals

Question

Rewrite the following integrals using the indicated order of integration and then evaluate the resulting integral. $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{\sqrt{1-x^{2}}} d y d z d x \text { in the order } d z d y d x$$

Step-by-Step Solution

Answer

Based on the given integral, we first analyze the region it represents in the \(xyz\)-plane. We find that this region is a sphere with radius 1 centered at the origin. Then, we rearrange the integral's order to \(dz\, dy\, dx\) and find the corresponding bounds for each variable. Finally, we evaluate the integral step by step, resulting in a value of \(\frac{1}{15}\).

Step 1: Analyze the given integral

We first need to understand the given integral and the region it describes. To do this, let's examine the limits for each variable. The bounds for \(z\) and \(y\) are the same which are from \(0\) to \(\sqrt{1-x^2}\), and for \(x\), it's from \(0\) to \(1\). So the region lies in the first octant and it's bound by the equation \(z^2 + y^2 = 1-x^2\) or \(x^2 + y^2 + z^2 = 1\). This is the equation of a sphere with radius 1 and centered at the origin.

Step 2: Rearrange the integral order

We need to rearrange the given integral into the order \(dz\, dy\,dx\). To do this, we must find the bounds for \(z\), \(y\), and \(x\). First, we rewrite the equation as $$z = \sqrt{1-x^2-y^2}$$ Now let's find the limits for each variable. Since we are in the first octant, \(x\geq0, y\geq0, z\geq0\). From the equation, we can express y as: $$y^2 \leq 1-x^2$$ $$0 \leq y \leq \sqrt{1-x^2}$$ For x, we have: $$x^2 \leq 1$$ $$0 \leq x \leq 1$$ From this, we can rearrange the integral order of bounds as follows: $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{\sqrt{1-x^{2}-y^{2}}} d z d y d x$$

Step 3: Evaluate the integral

Now, we can evaluate the given integral step by step starting with \(z\): $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \left[ z\right]_{0}^{\sqrt{1-x^{2}-y^{2}}} d y d x = \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \sqrt{1-x^{2}-y^{2}} d y d x$$ Next, we evaluate the integral with respect to \(y\): $$\int_{0}^{1} \left[\frac{1}{3} (1-x^2)^{3/2} \right]_{0}^{\sqrt{1-x^{2}}} d x = \int_{0}^{1} \frac{1}{3} (1-x^2)^{3/2} d x$$ Finally, we evaluate the integral with respect to \(x\) using the substitution method: Let \(u = 1-x^2 \Rightarrow du = -2x dx\), and the new bounds for \(u\) are 1 for \(x=0\) and 0 for \(x=1\): $$-\frac{1}{6} \int_{1}^{0} u^{3/2} d u = \frac{1}{6} \int_{0}^{1} u^{3/2} d u = \frac{1}{6} \left[\frac{2}{5} u^{5/2} \right]_{0}^{1} = \frac{1}{6} \cdot \frac{2}{5} = \frac{1}{15}$$ Thus, the value of the integral is \(\frac{1}{15}\).