This article covers the concept of **multi-variable calculus** and the aim is to understand the **double integrals,** how to **evaluate** and **simplify** them, and how they can be used to calculate the **volume** bounded by two **surfaces** or the area of a plane region over a **general region.** We will also learn how to simplify the **Integral calculations** by changing the **order** of integration and recognize if the functions of two **variables** are integrate-able over a region.

Volume is a **scalar** quantity defining the portion of three-dimensional **space** surrounded by a **closed** surface. Integrating a **curve** for any given limit gives us the **volume** that lies under the **curve** between the limits. Similarly, if the solid contains 2 **variables** in its equation, a double integral will be used to calculate its **volume.** We will first **integrate** the $dy$ with the given **limits** of $y$ and then **integrate** again the obtained result with $dx$ and this time with $x$ **limits.** Depending upon the **equation** of the **solid,** the **order** can be changed to make the **calculation** simpler, and $dx$ can be integrated before $dy$ and **vice versa.**

## Expert Answer

Given the **equation** of the solid is $z = 6-y$.

**Limits** are given as:

$ 0< x \leq 3$

$ 0< y \leq 4$

**Formula** for finding the volume is given as:

\[ V = \underset{y}{\int} \underset{x}{\int} z dydx \]

Now **inserting** the limits of $x$ and $y$ and **expression** $z$ in the **equation** and solving for $V$:

\[ V = \int_0^3 \int_0^4 (6 – y) dydx \]

Solving the internal **integral** $dy$ first:

\[V = \int_0^3 \left[ 6y – \dfrac{y^2}{2} \right]_0^4 dx\]

Now inserting the limits of $dy$ and subtracting the **expression** of the **upper limit** with an expression of the **lower limit:**

\[ V = \int_0^3 \left[ 6(4) – \dfrac{(4)^2}{2} \right] – \left[ 6(0) – \dfrac{(0)^2}{2} \right] dx \]

\[ V = \int_0^3 \left[ 24 – \dfrac{16}{2} \right] dx \]

\[ V = \int_0^3 \left[ 24 – 8 \right] dx \]

\[ V = \int_0^3 16 dx \]

Now that the only **outer integral** is left, solving for $dx$ to find the final answer of $V$.

\[ V = \int_0^3 16 dx \]

\[ V = [16x]_0^3 \]

Inserting the **limits** and **subtracting:**

\[ V = [16(3) – 16(0)] \]

\[ V = 48 \]

## Numerical Answer:

The volume of the **solid** using **double integral** is $V = 48$.

## Example

The **equation** of the solid is: $z = x – 1$ with limits $0< x \leq 2$ and $ 0< y \leq 4$. Finds its **volume.**

Applying the **formula:**

\[ V = \underset{y}{\int} \underset{x}{\int} z dydx \]

Inserting the **limits** and $z$:

\[ V = \int_0^2 \int_0^4 (x – 1) dydx \]

Solving $dy$ first:

\[ V = \int_0^2 \left[ xy – y \right]_0^4 dx \]

\[ V = \int_0^2 \left[ x(4) – 4 \right] – \left[ x(0) – 0 \right] dx \]

\[V = \int_0^2 4x -4 dx\]

Solving for $dx$ to obtain the **final answer** of $V$.

\[V = \left[ \dfrac{4x^2}{2} – 4x \right]_0^2 \]

Inserting the **limits** and **subtracting:**

\[ V = 2(2)^2 – 4 \]

\[ V = 4 \]