Question

Find the volume of the solid whose base is bounded by the curve y = 2x^3, the line x = 2 and the line y = 0, and the cross sections perpendicular to the y-axis are equilateral triangles.
What does one do to solve this? I know the formula used is the integral of (s^2)(3^.5)/4, but how does one find the parameters and is there something else that is missing? I can't seem to get the answer of (8/5)(3^.5)

Answer 1

Got you brother, just gimme a sec

Answer 2

Here, you are using the formula
\(\int_{c}^{d}A\left(y\right)dy\)
becuase it says 'cross sections perpendicular to the y-axis'

Now, the cross-sections are equilateral triangles. In order to find the volume bounded by the region, we must realize that Volume is the integral of the area or \(V=\int_{ }^{ }A\) Using this ideology, we must find the area of the curve. The area of an equilateral triangle is \(\frac{\sqrt{3}}{4}s^{2}\) I converted them into inverse functions to make it easier \(\int_{0}^{16}\left(2-\left(\frac{x}{2}\right)^{\frac{1}{3}}\right)dx\)

Answer 3

Wait why is that the integral? Where did the 2 and the (x/2)^1/3 come from?

Answer 4

Nah I took the inverses of all the functions and instead of integrating with respect to y I integrated with respect to x. It's an integration technique.

Answer 5

I got the answer as \(8\frac{\sqrt{3}}{4}\), is it correct?

Answer 6

Almost, but it's 8/5, not 8/4

Answer 7

Oof sorry I typed the integral wrong :/

Answer 8

Yeah I got that, lemme rephrase and retype

Answer 9

Using the inverse integration technique:

\(\frac{\sqrt{3}}{4}\int_{0}^{16}\left(2-\left(\frac{x}{2}\right)^{\frac{1}{3}}\right)^{2}dx = \frac{8\sqrt{3}}{5}\)

Answer 10

If you notice what I did, I just took the inverse of the functions and did the problem the same way, and integrated with respect to x this time. It's a bit easier because it makes it easier to find the s value to find the area of the equilateral triangle

Answer 11

So then in that, s=2-(x/2)^1/3
Plug that into the area of an equilateral triangle and then integrate with respect to x.

Answer 12

Wait, so my question is when can you use inverses with these sort of things? I've never done something like this before. Also, I'm still confused as to how the 2 came into play and the reasoning for it. Did you take the inverse of x = 2 and change that as y = 2? And even then, why are you able to keep the parameters of 0 and 16 if they are both in respect to y?

Answer 13

So in my class I learned about inverse integration techniques as a bonus topic when it came to integrating with respect to y.

Take a look at it graphically and maybe view a few videos.

Look at this desmos thing I explained it for you:

https://www.desmos.com/calculator/9qofmgq5vm

Answer 14

The parameters are the same because they are inverses. If you remember, inverses are where (a,b) change to (b,a). This lets you integrate with respect to x.

But you can also do integration with respect to y. I didn't use it here because I'm now so used to using inverse integration

Answer 15

Hey, thanks for the help. I really appreciate it! I think I learned a lot more this way than I did in class.

Answer 16

No problem man!

Answer 17

I think you'll get into more integration techniques when you're in second semester taking BC. My teacher told us this now but I'm not sure why. Still quite helpful!

Answer 18

Yeah, all I'm aware of is learning about integration by parts and taking the limit of an integral.

Answer 19

For Calc BC

Answer 20

Wait you're already doing that?! Wow all we did was u-substitution, but we'll do integration by parts and all those goodies next semester

Answer 21

Oh no, I didn't do any of those yet. That was what we're expected to do in BC. Also, I learned how to use u sub, but we were actually taught to use it on only very few integrals; supposedly our teacher only uses u sub for 1 out of 1000 integrals.