Question

Evaluate the following integrals. $$\int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{\sqrt{1+x^{2}+z^{2}}} d y d x d z$$

Short answer

Answer: The final value of the triple integral is 27.

Step-by-step solution

Identify the integrand and variables

The triple integral given is $$\int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{\sqrt{1+x^{2}+z^{2}}} d y d x d z$$ The integrand is 1, and we are integrating with respect to y, x, and z, in that order.

Evaluate innermost integral

First, we will integrate with respect to y: $$\int_{0}^{\sqrt{1+x^{2}+z^{2}}} dy$$ As the integrand is 1, the integral is simply y, which we evaluate between the limits of 0 and \(\sqrt{1+x^{2}+z^{2}}\). We get: $$[\sqrt{1+x^{2}+z^{2}} - 0] = \sqrt{1+x^{2}+z^{2}}$$

Evaluate the next integral

Now, we will integrate the result obtained in Step 2 with respect to x: $$\int_{0}^{\sqrt{9-z^2}} \sqrt{1+x^2+z^2} dx$$ We can proceed to integrate it using a substitution: let \(x=\sqrt{9-z^2}\sin{u}\). Then, \(dx=\sqrt{9-z^2}\cos{u}du\) and the integral will be evaluated from 0 to \(\pi/2\). Substitute and perform the integration with respect to u: $$\int_{0}^{\frac{\pi}{2}} \sqrt{1+(\sqrt{9-z^2}\sin{u})^2+z^2}(\sqrt{9-z^2}\cos{u}) du$$ The integral simplifies to: $$\int_{0}^{\frac{\pi}{2}} \sqrt{9-z^2}\sqrt{1+z^2} \cos{u} du$$ Now we can evaluate the integral: $$\sqrt{9-z^2}\sqrt{1+z^2} [\sin{u}]_{0}^{\frac{\pi}{2}} = \sqrt{9-z^2}\cdot\sqrt{1+z^2} (1)$$

Evaluate the outermost integral

Finally, we will integrate the result obtained in Step 3 with respect to z: $$\int_{0}^{3} \sqrt{9-z^2}\cdot\sqrt{1+z^2} dz$$ To evaluate this integral, we will use another substitution: let \(z=3\cos{v}\). Then, \(dz=-3\sin{v}dv\) and the integral will be evaluated between \(\pi/2\) and \(0\). Substitute and perform the integration with respect to v: $$\int_{\frac{\pi}{2}}^{0} \sqrt{9-(3\cos{v})^2}\cdot\sqrt{1+(3\cos{v})^2} (-3\sin{v}) dv$$ The integral simplifies to: $$-\int_{\frac{\pi}{2}}^{0} 9\sin^2{v}\cdot3\cos{v} dv$$ Now we can evaluate the integral: $$-27\int_{\frac{\pi}{2}}^{0} (\cos{v} - \cos^3{v}) dv = -27([\sin{v} - \frac{1}{4}\sin^3{v}]_{\frac{\pi}{2}}^0)$$ After evaluating the function between the limits, we get: $$-27[(0-\frac{1}{4}(0)^3) - (\cos{\frac{\pi}{2}} - \frac{1}{4}\sin^3{\frac{\pi}{2}})] = -27[- 1]$$ Therefore, the final value of the triple integral is: $$27$$.

Key concepts

Integration Techniques

Triple integrals, like the one in this exercise, require a proper understanding of different integration techniques. Integration here is performed with respect to three variables, typically in a nested, stepwise fashion. The exercise exhibits an example where the variables are integrated sequentially—starting with the innermost integrand and proceeding outward. This method is essential for solving problems involving volumes and other spatial calculations. By simplifying the integrand step by step, it helps tremendously in solving complex integrals. It's crucial to follow a structured integration path: first evaluate the innermost integral, then progress to the next level, continuing this pattern until all variables have been integrated.

Substitution Method

The substitution method is a powerful integration technique that can simplify complex integrals. In this exercise, substitution is used twice, once while integrating with respect to x and another time with respect to z.

• For the integration with respect to x, the substitution was made using trigonometric substitution: specifically, letting \( x = \sqrt{9-z^2} \sin{u} \) which converted the integral into terms involving sinusoidal functions.
• For the outermost integral with respect to z, a similar trigonometric substitution was employed: \( z = 3 \cos{v} \).

Such substitutions are commonly used to simplify the integrals involving square roots and can convert a cumbersome integral into a more manageable form, often involving trigonometric identities. This typically transforms limits of integration to angles, which then can be manipulated using trigonometric properties, easing the integration process. Substitution not only helps in simplifying the mathematical expression but also aids in revealing symmetries or hidden properties of the problem.

Limits of Integration

Understanding the limits of integration is crucial for accurately solving integrals, especially when dealing with complex geometrical shapes or regions. This exercise demonstrates how each limit governs the extent of integration for each of the three axes.

• The innermost integral had limits from 0 to \( \sqrt{1 + x^2 + z^2} \), which describes a variable boundary dependent on the values of x and z.
• The next integral bounds x from 0 to \( \sqrt{9 - z^2} \), illustrating a semi-circular cross-section in the xz-plane.
• The final integral bounds z from 0 to 3, cleaning the integration region to a hemisphere-like shape along the z-axis.

When solving triple integrals, the limits often represent the boundaries and shapes of the solid region of interest. Making sure to correctly translate these geometric constraints into limits within the integrals is essential for obtaining the correct solution. Identifying these limits precisely often requires a solid understanding of the geometry related to the problem, allowing you to set up the integrals accurately and to understand the space over which the function is being integrated.