Use a double integral to find the volume of the solid shown in the figure.

Figure-1

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.

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:

$V = [16(3) – 16(0)]$

$V = 48$

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:

$V = 2(2)^2 – 4$

$V = 4$