Introduction

Welcome back! Today, we’ll begin to dive into the harder topics of Unit 8, namely finding volumes. We’ll be starting off simple though, as we’re only going to focus on volumes that have either square or rectangular cross sections. Now, how exactly do you find the volume of a solid with square or rectangular cross sections?

Finding Volume

The simplest way to think about finding volumes is to think about finding the area under a curve, except the curve is now the edge of a solid. This essentially means we are going to take the area under the curve, and multiply it by how far the solid extends into the third dimension at every single point on the curve. The way we’ll be doing this is by integrating the formula for the area of each cross section. This formula is actually really simple, as it just involves plugging in the function of the curve into the formula for area. 

Example

For example, the formula for the area of a square is given by $A=s^2$, where $s$ represents the length of the sides of the square. If we wanted to find the volume of a solid with square cross sections that was bounded by the $x$-axis, $y$-axis, the line $x=3$, and the graph of $f(x)=3x^2$, we would simply substitute $f(x)$ for $s$, and then integrate. In this case, we have our bounds set as $x=0,3$, so we would use the integral ${\int }_0^3(3x^2{)}^{2}dx$. This would then give us $437.4$. We can do the same thing for rectangles, but since the area of a rectangle is given by $A=lw$, where $l$ is length and $w$ is width, we would need to have our width defined already since our functions only give us length (or vice versa). Now that you have a basic understanding of how to find volumes, let’s go over to the practice section.

Practice

Answers

1) In this problem, we know that $y=3x^2$ is always greater than the x-axis ($y = 0$), and our bounds have been given too. Since our cross sections are squares, we’ll use the formula $A=s^2$, but replace $s$ with $3x^2-0$. Now all we have to do is set up the integral and solve.

Plugging in:

$V={\int }_0^2{\left(3x^2-0\right)}^{2}dx$

$V={\int }_0^2{\left(3x^2\right)}^{2}dx$

$V={\int }_0^2\left(9x^4\right)dx$

$V={\left[\frac{9x^5}{5}\right]}_0^2$

$V=\left[\frac{9(2{)}^5}{5}-\frac{9(0{)}^5}{5}\right]$

$V=\left[\frac{9(32)}{5}-\frac{9(0)}{5}\right]$

$V=\left[\frac{288}{5}-0\right]$

$V=\frac{288}{5}$

Therefore, the volume of the solid is $\frac{288}{5}$.

2) In this problem, our first issue is finding out what we need to integrate, as our bounds have already been given. We have to use the formula $A=hw$, where $h$ is the height and $w$ is the width. We can find the width by doing $\sqrt{x}-x^2$, since $x^2$ has smaller values on $[0,1]$. Our height is twice our width, so it becomes $2\left(\sqrt{x}-x^2\right)$. With this, we can plug everything into our formula and solve.

Plugging in:

$V={\int }_0^1\left(\sqrt{x}-x^2\right)\left(2\left(\sqrt{x}-x^2\right)\right)dx$

$V={\int }_0^1\left(\sqrt{x}-x^2\right)\left(2\sqrt{x}-2x^2\right)dx$

$V={\int }_0^1\left(2x-2x^{\frac{5}{2}}-2x^{\frac{5}{2}}+2x^4\right)dx$

$V={\int }_0^1\left(2x-4x^{\frac{5}{2}}+2x^4\right)dx$

$V={\left[x^2-\frac{8}{7}{x}^{\frac{7}{2}}+\frac{2x^5}{5}\right]}_0^1$

$V=\left[(1{)}^2-\frac{8}{7}(1{)}^{\frac{7}{2}}+\frac{2(1{)}^5}{5}\right]-\left[(0{)}^2-\frac{8}{7}(0{)}^{\frac{7}{2}}+\frac{2(0{)}^5}{5}\right]$

$V=\left[1-\frac{8}{7}(1)+\frac{2}{5}\right]-\left[0-0+0\right]$

$V=\left[\frac{35}{35}-\frac{40}{35}+\frac{14}{35}\right]$

$V=\frac{9}{35}$

Therefore, the volume of the solid is $\frac{9}{35}$.