Changing The Order of Integration in Triple Integrals

# Changing The Order of Integration in Triple Integrals

Suppose that we want to integrate the three variable real-valued function $w = f(x, y, z)$ over the region $E$ in $\mathbb{R}^3$. Then we will need to evaluate the triple integral $\iiint_E f(x, y, z) \: dV$ in terms of triple iterated integrals. There will be six different orders of evaluating the triple iterated integrals. We can:

• Integrate with respect to $z$ first, and then with respect to $x$ and then $y$ (Type 1 Region).
• Integrate with respect to $z$ first, and then with respect to $y$ and then $x$ (Type 1 Region).
• Integrate with respect to $x$ first, and then with respect to $y$ and then $z$ (Type 2 Region).
• Integrate with respect to $x$ first, and then with respect to $z$ and then $y$ (Type 2 Region).
• Integrate with respect to $y$ first, and then with respect to $x$ and then $z$ (Type 3 Region).
• Integrate with respect to $y$ first, and then with respect to $z$ and then $x$ (Type 3 Region).

Now it is important to be able to change the order of integration because in some problems, it may be difficult or even impossibly to evaluate a particular order of integration, but much easier to evaluate a different order of integration just like with evaluating double integrals.

Let's now look at some examples of changing the order of integration in triple integrals.

## Example 1

Rewrite $\iiint_E f(x, y, z) \: dV$ in all six orders of integration where $E$ is the solid bounded by the elliptic paraboloid $y = 4 - x^2 - 4z^2$ and the plane $xz$-plane.

It is first important to note that the $xz$ plane can be expressed as $y = 0$ in $\mathbb{R}^3$. Furthermore, the elliptic paraboloid $y = 4 - x^2 - 4z^2$ is parallel to the $y$-axis and opens negatively. The following is a graph of our region $E$.

Now let's start by setting up the triple iterated integrals as though $E$ is a type 1 region. In doing so, we want our values of $z$ to be inside this elliptic paraboloid. We need to find the boundaries of our elliptic paraboloid in terms of the variables $x$ and $y$. This can easily be done by taking $y = 4 - x^2 - 4z^2$ and isolating for $z$:

(1)
\begin{align} \quad 4z^2 = 4 - x^2 - y \\ \quad z^2 = 1 - \frac{x^2}{4} - \frac{y}{4} \\ \quad z = \pm \sqrt{1 - \frac{x^2}{4} - \frac{y}{4}} \end{align}

Therefore we want $z$ to be inside our paraboloid, so $-\sqrt{1 - \frac{x^2}{4} - \frac{y}{4}} ≤ z ≤ \sqrt{1 - \frac{x^2}{4} - \frac{y}{4}}$.

Now let's look at the "projection" or "shadow" of $E$ onto the $xy$ plane. It's not hard to see that we will get a negative parabola. This can be determined by noting that the projection/shadow of $E$ can be obtained by taking points inside the curve $4(0)^2 = 4 - x^2 - y$, that is $y = -x^2 + 4$, which is graphed below.

If we want to integrate with respect to $x$ secondly, then note that $-\sqrt{4 - y} ≤ x ≤ \sqrt{4 - y}$, and then $0 ≤ y ≤ 4$. Therefore one order for our triple iterated integrals is:

(2)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \int_0^4 \int_{-\sqrt{4 - y}}^{\sqrt{4 - y}} \int_{-\sqrt{1 - \frac{x^2}{4} - \frac{y}{4}}}^{\sqrt{1 - \frac{x^2}{4} - \frac{y}{4}}} f(x, y, z) \: dz \: dx \: dy \end{align}

If we want to integrate with respect to $y$ secondly, then we note that $0 ≤ y ≤ -x^2 + 4$ and $-2 ≤ x ≤ 2$. Therefore another order for our triple iterated integrals is:

(3)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \int_{-2}^{2} \int_{0}^{-x^2 + 4} \int_{-\sqrt{1 - \frac{x^2}{4} - \frac{y}{4}}}^{\sqrt{1 - \frac{x^2}{4} - \frac{y}{4}}} f(x, y, z) \: dz \: dy \: dx \end{align}

Now let's treat $E$ as a type 2 region, that is, integrating with respect to $x$ first. We will want to have $x$ be contained with our paraboloid. If we take the equation $y = 4 - x^2 - 4z^2$ and isolate for $x$ we get that:

(4)
\begin{align} \quad x^2 = 4 - y - 4z^2 \\ \quad x = \pm \sqrt{4 - y - 4z^2} \end{align}

Therefore we have that $- \sqrt{4 - y - 4z^2} ≤ x ≤ \sqrt{4 - y - 4z^2}$.

Once again, it is not hard to see that the region $E$ will cast a parabolic shadow onto the $yz$-plane. This equation of this parabola can be obtained by setting $x$ equal to $0$ in $y = 4 - x^2 - 4z^2$ to get $y = 4 - (0)^2 - 4z^2$ or simply $y = 4 - 4z^2$

If we want to integrate with respect to $y$ secondly, then we will have that $0 ≤ y ≤ 4 - 4z^2$ and $-1 ≤ z ≤ 1$, and so another order for our triple iterated integrals is:

(5)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \int_{-1}^{1} \int_0^{4 - 4z^2} \int_{-\sqrt{4 - y - 4z^2}}^{\sqrt{4 - y - 4z^2}} f(x, y, z) \: dx \: dy \: dz \end{align}

If we want to integrate with respect to $z$ secondly, then we will have that $-\sqrt{1 - \frac{y}{4}} ≤ z ≤ \sqrt{1 - \frac{y}{4}}$ and $0 ≤ y ≤ 4$, and so another order for our triple iterated integrals is:

(6)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \int_0^4 \int_{-\sqrt{1 - \frac{y}{4}}}^{\sqrt{1 - \frac{y}{4}}} \int_{-\sqrt{4 - y - 4z^2}}^{\sqrt{4 - y - 4z^2}} f(x, y, z) \: dx \: dz \: dy \end{align}

Now let's three $E$ as a type 3 region, that is, integrating with respect to $y$ first. We want to have $y$ contained inside our paraboloid but also above the the plane $y = 0$. Thus $0 ≤ y ≤ 4 - x^2 - 4z^2$.

Now it shouldn't be too difficult to see that the projection/shadow of the region $E$ onto the $xz$ plane forms an ellipse. This can easily be seen by setting $y$ equal to $0$ in the equation $y = 4 - x^2 - 4z^2$ to get $0 = 4 - x^2 - 4z^2$, that is $x^2 + 4z^2 = 4$.

If we want to integrate with respect to $x$ secondly, then we will have that $-\sqrt{4 - 4z^2} ≤ x ≤ \sqrt{4 - 4z^2}$ and $-1 ≤ z ≤ 1$. So another order for our triple iterated integrals is:

(7)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \int_{-1}^{1} \int_{-\sqrt{4 - 4z^2}}^{\sqrt{4 - 4z^2}} \int_0^{4 - x^2 - 4z^2} f(x, y, z) \: dy \: dx \: dz \end{align}

If we want to integrate with respect to $z$ secondly, then we will have that $- \sqrt{1 - \frac{x^2}{4}} ≤ z ≤ \sqrt{1 - \frac{x^2}{4}}$ and $-2 ≤ x ≤ 2$. So the last order for our triple iterated integrals is:

(8)
\begin{align} \quad \iiint_E f(x, y, z) \: dV = \int_{-2}^{2} \int_{- \sqrt{1 - \frac{x^2}{4}}}^{\sqrt{1 - \frac{x^2}{4}}}\int_0^{4 - x^2 - 4z^2} f(x, y, z) \: dy \: dz \: dx \end{align}