The base of a solid is a circle of radius a, and its vertical cross sections are equilateral triangles. Find the radius of the circle if the volume of the solid is 10 cubic meters.

Question

mathslover · Answer

I think calculus would be needed here

sw050399 · Answer

You are correct. This is a practice problem from my Calculus I class that I didn't really understand. 

If you have anything to contribute, it would be greatly appreciated :)

mathslover · Answer

it is clear that circle is in the XY plane. So the integration would be along X or Y axis. Well it does not matter.

sw050399 · Answer

Okay, I get that part.

mathslover · Answer

Well you can put the triangles perpendicular to the X-axis

mathslover · Answer

I am going to use x - axis here.

sw050399 · Answer

Not a problem.

mathslover · Answer

Let dA be the area of the triangles . 

So volume = $\int {dA dx}$

mathslover · Answer

calculate dA , $dA = \cfrac{1}{2} b\times h$

sw050399 · Answer

Because it's an equilateral triangle, would you have to involve $$((s^2)(\sqrt{3}))/4 $$ somehow?

mathslover · Answer

well .. do you agree that base = 2y

sw050399 · Answer

Yes. It's an equilateral triangle and because it's perpendicular to the x axis all sides are equal to 2y.

mathslover · Answer

Yep! $h = y \tan 60$ or $h = y\sqrt{2}$

mathslover · Answer

So $dA = ?$ can u tell me?

mathslover · Answer

You have to just apply : 1/2 bh formula.. you have b = 2y and h = y$\sqrt{2}$

sw050399 · Answer

How did you get sqrt2? Isn't it sqrt3?

mathslover · Answer

h = y tan 60 ... agreed?

mathslover · Answer

Yeah!

mathslover · Answer

sqrt{3} sorry

mathslover · Answer

So we have $dA = ?$ ..

sw050399 · Answer

$$dA = (1/2)(2y)(y \sqrt{3})$$
$$= y^2 \sqrt{3}$$

mathslover · Answer

yes!

mathslover · Answer

Find $\int y^2 \sqrt{3} dx$

mathslover · Answer

you can shorten your work : 

$y^2 = r^2 -x^2$ 

put that there on the place of $y^2$

sw050399 · Answer

$$\int\limits(r^2 - x^2)(\sqrt{3})dx$$
$$((r^3)/3 - (x^3)/3)*((6\sqrt{3})/3)$$
$$(1/3)(r^3 - x^3)(2\sqrt{3}) + C$$

sw050399 · Answer

I forgot the + C in the 2nd line...

mathslover · Answer

I think it will be : $\sqrt{3}(r^2 x - \cfrac{x^3}{3}) + C$

mathslover · Answer

check ur method again!

mathslover · Answer

$\int (r^2-x^2)\sqrt{3} dx$
$\sqrt{3} \int (r^2-x^2)dx$
$\sqrt{3} [\int r^2 dx - \int x^2 dx]$
$\sqrt{3} [r^2x - \cfrac{x^3}{3}]+C$ 

got it?

sw050399 · Answer

Yes. I actually got right before you posted the work... 

I just forgot that we were integrating with respect to x.

mathslover · Answer

Now evaluate it over the range : -r to r ...

mathslover · Answer

note: evalutate it in X

mathslover · Answer

then equate it to 10 unit^3

find r

sw050399 · Answer

$$r = \sqrt{30/4\sqrt{3}} - C$$

mathslover · Answer

would it be cube root?

sw050399 · Answer

Yes, I just forgot to type that part.

mathslover · Answer

ok! and I think there would be no -C

sw050399 · Answer

Right, because the the other part is just a constant and adding some some that stays the same all the time isn't going to do anything.

mathslover · Answer

So you have your answer .. solve that by using calculator or let it be like that only.

mathslover · Answer

though @hartnn would check out the whole method ..

sw050399 · Answer

Thank you for your help!! I'll try that answer and see how it goes. I think we're right, but my teacher's kinda crazy, so I'll let you know.

Thank you again!! :)

mathslover · Answer

You're welcome and Best of Luck!