Introduction

Welcome back! Today’s topic is a continuation of the last topic, except our cross sections now take the shapes of triangles and semicircles. These volumes are a little trickier to find, as getting the areas of triangles and semicircles is slightly more difficult as there are multiple places where the function can be substituted.

Finding Volumes Continued

As you probably know, the area of a triangle can be given by $A=\frac{1}{2}bh$, where $b$ is the length of the triangle’s base and $h$ is the height of the triangle.

For a semicircle, it’s simply the area of a circle cut in half, so $A=\frac{\pi }{2}{r}^2$, where $r$ is the radius of the circle. However, when we are given a function for semicircles in AP Calculus, they are always considered to be the diameter. Therefore, our formula becomes $A=\frac{\pi }{2}{\left(\frac{\text{D}}{2}\right)}^2$. Simplifying this gives us $A=\frac{\pi }{2}\left(\frac{D^2}{4}\right)$, which can be further simplified to $A=\frac{\pi }{8}{D}^2$, where $D$ is the given function. For the base, it will usually either be given or it will be a specific distance that needs to be calculated.

If you are given an equilateral triangle for your cross sections, the formula $A=\frac{\sqrt{3}}{4}{s}^2$ can be used. In case you’re still confused, we’ll do an example together. Again, depending on whether your solids are generated perpendicular to the $x$- or $y$-axis, you will have to integrate with respect to either $x$ or $y$.

Example

Let’s look at the following problem:

1) The base of a solid is the region bounded by the curve $y=x^2$, the $x$-axis, the $y$-axis, and the line $x=3$. The solid’s cross sections are isosceles triangles that are perpendicular to the $y$-axis, and their height is given by the value of $y$ when $x=2$. Find the volume of the solid.

Now this problem looks extremely complex, but that’s mostly because the parts we need have already been defined. Our region is neatly bounded on both sides at $x=0,3$, so we have the bounds of the integral already. Next, all we need to do is find $y$ when $x=2$, which when plugged into $y=x^2$ comes out to be $4$. That leaves the base, which is where we plug in our function.

Doing everything gives us $V=\frac{1}{2}(4){\int }_0^3{x}^{2}dx$. The $\frac{1}{2}$ and the $4$ come from the $\frac{1}{2}$ in $\frac{1}{2}bh$, and the $4$ is our $b$, which we can take out of the integral since it is a constant. 

Taking the integral gives us $V=2{\left[\frac{x^3}{3}\right]}_0^3$, which when evaluated simplifies to $V=2[9-0]$, which become $V=18$.

See? That wasn’t so bad was it? If it was, I’d say review a lot, as finding volumes makes up the entire last half of unit 8. Anyways, practice makes perfect, so onto the practice section we go. This one will just be a quick problem on semicircles, as we already covered how to do triangles. As a quick reminder, the process for equilateral triangle cross sections is the same as what we just did, except you will use the formula $A=\frac{\sqrt{3}}{4}{s}^2$ and plug your function into $s$.

Practice

Answers

Setup:

$V=\frac{\pi }{8}{\int }_0^4{\left(4x^2+10\right)}^{2}dx$

Solving:

$V=\frac{\pi }{8}{\int }_0^4(4x^2+10{)}^{2}dx$

$V=\frac{\pi }{8}{\int }_0^4(16x^4+80x^2+100)dx$

$V=\frac{\pi }{8}{\left[\frac{16x^5}{5}+\frac{80x^3}{3}+100x\right]}_0^4$

$V=\frac{\pi }{8}\left[\frac{16(4{)}^5}{5}+\frac{80(4{)}^3}{3}+100(4)\right]$

$V=\frac{\pi }{8}\left[\frac{16(1024)}{5}+\frac{80(64)}{3}+400\right]$

$V=\frac{\pi }{8}\left[\frac{16384}{5}+\frac{5120}{3}+400\right]$

$V=\frac{\pi }{8}\left[\frac{49152}{15}+\frac{25600}{15}+\frac{6000}{15}\right]$

$V=\frac{\pi }{8}\cdot (\frac{80752}{15})$

$V=\frac{10094\pi }{15}$

Therefore, the volume of the solid is $\frac{10094\pi }{15}$.