Problem 25 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the given triple integral: \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{\sqrt{1-x^2}} d z d y d x\). Solution: The value of the triple integral is \(\frac{2}{3}\).

Step 1: Apply the limits of integration for z.

Integrate the function with respect to z along the given limits of integration (0 to \(\sqrt{1-x^2}\)). Since we are just integrating 1 (the function is 1), the result will simply be the difference between the upper and lower limits: $$\int_{0}^{\sqrt{1-x^{2}}} d z = \sqrt{1-x^2} - 0 = \sqrt{1-x^2}$$ Now our integral looks like this: $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \sqrt{1-x^2} dy dx$$

Step 2: Apply the limits of integration for y.

Integrate the function (\(\sqrt{1-x^2}\)) with respect to y along the given limits of integration (0 to \(\sqrt{1-x^2}\)): $$\int_{0}^{\sqrt{1-x^{2}}} \sqrt{1-x^2} dy = \left[\sqrt{1-x^2}y\right]_0^{\sqrt{1-x^2}} = (\sqrt{1-x^2})(\sqrt{1-x^2}) - (\sqrt{1-x^2})(0) = 1 - x^2$$ Now our integral looks like this: $$\int_{0}^{1} (1 - x^2) dx$$

Step 3: Apply the limits of integration for x.

Integrate the function (\(1 - x^2\)) with respect to x along the given limits of integration (0 to 1): $$\int_{0}^{1} (1 - x^2) dx = \left[x - \frac{1}{3}x^3\right]_0^1 = (1 - \frac{1}{3}(1)^3) - (0 - \frac{1}{3}(0)^3) = 1 - \frac{1}{3} = \frac{2}{3}$$ The value of the triple integral is \(\frac{2}{3}\).