I've tried this a couple times now, no luck: "Let the base of a solid be the first quadrant plane region bounded by y= 1- x^2/25, the x-axis, and the y-axis. Suppose that cross sections perpendicular to the x-axis are equilateral triangles sitting on the base. Find the volume of the solid."

I've attached screenshots of my work. Any help would be appreciated! :)

Question

I've tried this a couple times now, no luck: "Let the base of a solid be the first quadrant plane region bounded by y= 1- x^2/25, the x-axis, and the y-axis. Suppose that cross sections perpendicular to the  x-axis are equilateral triangles sitting on the base. Find the volume of the solid." 

I've attached screenshots of my work. Any help would be appreciated! :)

anonymous · Answer

attachment

anonymous · Answer

attachment

ganeshie8 · Answer

Area of equilateral triangle of side $s$        =  $\large  \frac{\sqrt{3}}{4} s^2$

ganeshie8 · Answer

right ?

ganeshie8 · Answer

and, $\large   s =   y =  1- \frac{x^2}{25} $

ganeshie8 · Answer

plug it  and integrate between 0 and 5 ?

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=%5Cint+%5Climits_0%5E5++sqrt%283%29%2F4+%281-x%5E2%2F25%29%5E2+dx

anonymous · Answer

Ah, I knew I was over-complicating it. :P Thank you. I'll try it out now.

anonymous · Answer

The volume of an object with cross-sections perpendicular to the x-axis that are equilateral triangles is expressed in the equation:
$$\frac{ \sqrt{3} }{ 4 }\int\limits_{a}^{b}f(x)^{2}dx$$
Hope this helps!

ganeshie8 · Answer

Nice drawings btw :)

anonymous · Answer

Thanks! :) My first time in school, I majored in design. I have to be careful, or I find myself drawing more than calculating.

anonymous · Answer

And thank you also, wotseit. I wish my book had provided that.

ganeshie8 · Answer

cool :) being able to interpret correctly and sketching is always your friend !

anonymous · Answer

It's strange that it didn't provide it... Glad I could help! Best of luck :)

ganeshie8 · Answer

I see nothing wrong in ur method,
you're just finding the height and trying to use 1/2ab formula right ?

ganeshie8 · Answer

you should get the same answer

anonymous · Answer

It would take far too long. It's better that he learns how to do it faster, especially when it shows up on a test.

ganeshie8 · Answer

o_o

ganeshie8 · Answer

In your first attachment, you made a mistake in simplification :

$\large  b =  \frac{\sqrt{3}}{2}  \left(1-\frac{x^2}{25}\right)$

ganeshie8 · Answer

^thats the height

ganeshie8 · Answer

Once u fix that,  since you knw the side =  $\large      1-\frac{x^2}{25}$, you can write out an expression for area element :

$\large      A =   \frac{1}{2}  \times     \frac{\sqrt{3}}{2}  \left(1-\frac{x^2}{25}\right)\times    \left(1-\frac{x^2}{25}\right)$

,

ganeshie8 · Answer

simplify, it gives the same integrand