Problem 14 - University Calculus: Early Transcendentals

Question

Evaluate the integrals in Exercises \(7-20\). $$\int_{0}^{2} \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} \int_{0}^{2 x+y} d z d x d y$$

Step-by-Step Solution

Answer

The evaluated integral is \(\frac{16}{3}\).

Step 1: Integrate with respect to z

First, evaluate the innermost integral with respect to the variable \(z\). The limits of integration for \(z\) are from \(0\) to \(2x + y\). Since the integrand is just 1, the integral becomes simply \(z\) evaluated from \(0\) to \(2x + y\). This results in: \[ \int_{0}^{2x+y} 1 \, dz = [z]_0^{2x+y} = 2x + y.\] Replace the innermost integral with \(2x + y\).

Step 2: Integrate with respect to x

Next, integrate the result from Step 1 with respect to \(x\) from \(-\sqrt{4-y^2}\) to \(\sqrt{4-y^2}\). The integral is:\[ \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} (2x+y) \, dx.\]Separate into two integrals:\[ 2\int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} x \, dx + y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} 1 \, dx.\]Calculate each integral. The first is:\[ 2\left[ \frac{x^2}{2} \right]_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} = 2\left[ \frac{(\sqrt{4-y^2})^2 - (-\sqrt{4-y^2})^2}{2} \right] = 0,\]since the terms cancel each other. The second integral is:\[ y \left[ x \right]_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} = y[\sqrt{4-y^2} - (-\sqrt{4-y^2})] = 2y\sqrt{4-y^2}.\]The result of integrating with respect to \(x\) is \(2y\sqrt{4-y^2}\).

Step 3: Integrate with respect to y

Finally, integrate the result from Step 2 with respect to \(y\) from 0 to 2:\[ \int_{0}^{2} 2y\sqrt{4-y^2} \, dy.\]This is best solved using the substitution method. Let \(u = 4 - y^2\), then \(du = -2y \, dy\) or \(ydy = -\frac{du}{2}\). Substituting, the limits change: when \(y = 0\), \(u = 4\); when \(y = 2\), \(u = 0\).\[ \int_{4}^{0} -\sqrt{u} \, rac{du}{2} = \frac{1}{2} \int_{0}^{4} \sqrt{u} \, du,\]where the negative sign reverses the limits:\[ \frac{1}{2} \left[ \frac{2u^{3/2}}{3} \right]_{0}^{4} = \frac{1}{3} [2(4)^{3/2} - 0] = \frac{1}{3} \times 2 \times 8 = \frac{16}{3}.\]This is the final evaluated integral.

Key Concepts

Triple integrals

Triple integrals are a way of integrating in three dimensions. They are essential when working with volumes or when integrating over three-dimensional spaces.
In simpler terms, a triple integral allows you to calculate the accumulation of a function over a 3D region.
When evaluating a triple integral, it’s helpful to understand that it comprises three nested single integrals. Each integral evaluates over a specific variable: typically, these are the variables \(x\), \(y\), and \(z\).
For a problem like the one given, you integrate successively—first with respect to \(z\), then \(x\), and finally \(y\). By integrating one variable at a time, you simplify the problem into manageable chunks. This process is known as iterated integration.

Substitution method

The substitution method is a powerful technique used to simplify the process of integration by changing variables.
This method is especially useful for complicated functions, like in the last step of the original problem, where you transform the integral to a simpler form.
The key idea is to replace a complex part of the function with a new variable, \(u\). For instance, in the given exercise, \(u = 4 - y^2\) was chosen to simplify the integral of \(2y\sqrt{4-y^2}\).
After defining \(u\), it's necessary to find \(du\), the derivative of \(u\) with respect to \(y\): \(du = -2y \, dy\). This substitution turns the difficult integral into one that is much simpler to solve.
Remember to also change the limits of integration when you change variables, as they need to correspond to the new variable \(u\). This ensures that calculations remain accurate and straightforward.

Integration limits

Integration limits define the range over which integration occurs. They are crucial as they determine the extent of the area, volume, or space you consider when integrating.
In the given triple integral exercise, each of the variables \(z\), \(x\), and \(y\) has a set of integration limits. These limits can depend on other variables; for example, the limits for \(z\) are based on \(x\) and \(y\).
The integration limits for \(z\) are from 0 to \(2x + y\). As you proceed to integrate with respect to \(x\), \(x\)'s limits are defined by \(-\sqrt{4-y^2}\) to \(\sqrt{4-y^2}\), which creates symmetry around the \(y\)-axis. Finally, as you integrate with respect to \(y\), the limits are from 0 to 2.
Intricately understanding these limits will guide how you evaluate each integral iteratively, ensuring all calculations remain within the intended bounds.

Iterated integration

Iterated integration refers to the step-by-step process of evaluating multiple integral calculations in sequence. In a triple integral, you perform iterated integration three times—one for each dimension.
You start with the innermost integral, as was done with \(z\) in the given exercise. This approach simplifies the problem by allowing you to deal with one variable at a time.
After resolving the \(z\)-integral, you move to the \(x\)-integral, and finally, the \(y\)-integral. Each step uses the result from the previous step, and this systematic approach prevents error and confusion during long calculations.
Key to mastering iterated integration is keeping track of the function and its limits, ensuring that each step logically follows from the last. By practicing iterated integration, solving complex multi-variable problems gradually becomes more intuitive.