A solid with its base as region R has equilateraltriangle-crosssections perpendicular to the x-axis

[crisp forge]

So I understand A, and could probably do C when I get to it, but B eludes me. My logic for A is that the points of intersection are 1 and 3 so by finding -x^2+4x-3 and then integral of Fb - Fa I get 4/3. I assume that is right, if it isn't please let me know. Regardless, for part B, I'm confused. I know the area of equilateral triangle is (Sqrt(3)/4 )* s^2 but I don't know what to do with that info, am I supposed to plug in the -x^2+4x-3 for s and then integrate from 1 to 3? Or am I supposed to do something else? I get (4sqrt(3))/15. Is this right?

[tawny heron]

Your method for (a) is right

Your method for (b) is right too

(-x^2 + 4x - 3) measures the height along R, and each of those heights is the base of an equilateral triangle.

To simplify what I'm going to say, let k(x) = -x^2 + 4x - 3.
If the question had instead said that it had rectangular cross sections with height 4, you would integrate 4*k(x).
If the question had instead said that it had square cross sections, you would integrate (k(x))^2.

[crisp forge]

ah, and just checking, for c should it be the integral from 1 to 3 (pi((f(x)^2)-(g(x)^2)) dx)

[tawny heron]

You got it.

[crisp forge]

alright, so everything is good?

[tawny heron]

Good to notice that you need the squares before you take the difference.

[crisp forge]

thanks

[tawny heron]

Yeah, it looks like everything is good.

[crisp forge]

👍

[tawny heron]

I wish I had something insightful to add, but you seem to know this.

[crisp forge]

oh no, I learned it a while ago, I was going off memory and was really unsure, so having clarification is still insanely helpful

Thank you

.solved

[velvet panther]

Solved

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