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Chapter 16: Problem 41

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

Short Answer

Expert verified
Answer: The value of the triple integral is $\frac{2}{3}$.

Step by step solution

01

Integrate with respect to z

Since there are no functions inside the integral, when integrating with respect to z, simply take the difference between the limits of z and multiply it by 1. $$\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2}} dz dy dx = \int_0^1 \int_0^{\sqrt{1-x^2}} (\sqrt{1-x^2} - 0) dy dx$$ The new integral after integrating with respect to z becomes: $$\int_0^1 \int_0^{\sqrt{1-x^2}} \sqrt{1-x^2} dy dx$$
02

Integrate with respect to y

Now that we have integrated with respect to z, we will integrate with respect to y. Notice that the integrand does not depend on y, so once again, we just find the difference between the limits of y and multiply by the integrand. $$\int_0^1 \int_0^{\sqrt{1-x^2}} \sqrt{1-x^2} dy dx = \int_0^1 (\sqrt{1-x^2} - 0) \cdot \sqrt{1-x^2} dx$$ The new integral after integrating with respect to y becomes: $$\int_0^1 (\sqrt{1-x^2})^2 dx$$
03

Simplify the Integral

Before integrating with respect to x, let's simplify the current integrand: $$(\sqrt{1-x^2})^2 = 1 - x^2$$ So now, we have: $$\int_0^1 (1 - x^2) dx$$
04

Integrate with respect to x

Finally, integrate with respect to x: $$\int_0^1 (1 - x^2) dx = \left[x - \frac{x^3}{3}\right]_0^1$$
05

Evaluate the Integral

Evaluate this expression: $$\left[x - \frac{x^3}{3}\right]_0^1 = (1 - \frac{1}{3}) - (0 - \frac{0}{3}) = \frac{2}{3}$$ Therefore, the value of the triple integral is: $$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{0}^{\sqrt{1-x^{2}}} d z d y d x = \frac{2}{3}$$

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