Question

The region bounded by the graph of y = 3x-x^2 and the x-axis is the base of a solid. Each cross section
perpendicular to the x-axis is an equilateral triangle. What is the volume of this solid?

Answer 1

i got about 3.507.

Answer 2

what is the volume of an equilat tri?

Answer 3

\[\int_{a}^{b}Area.equilat.tri\ wrt\ 3x-x^2\]

Answer 4

side^2 sin(60)/2 is the area of that tri if I remember it right

Answer 5

but we need to relate side to height to use the equation

Answer 6

s^2 + h^2 = (2s)^2

h^2 = 4s^2 - s^2

h^2 = 3s^2

s = h/sqrt(3) right?

whre h = f(x)

Answer 7

yes

Answer 8

i did root 3 divided by 4 and multiplied by integral from 0 to 3 with (3x-x^2)^2 as my function

Answer 9

are of tri = h^2/3 sin(60)
= sin(60)/3*2 (f(x))^2

out limits of integration are then the roots of f(x);

3x-x^2 = 0

x(3-x) = 0 when x=0 and x=3

\[\frac{sin(60)}{6}\int_{0}^{3}(3x-x^2)^2\ dx \]

looks about right to me for a setup

Answer 10

mighta got the denomnator off tho

Answer 11

\[\frac{s^2}{2}sin(60)\to \frac{h^2}{3^2*2}sin(60)\to \frac{\sqrt{3}}{36}[f(x)]^2\]

Answer 12

9x^2-6x^3+x^4 --> 3x^3 -3x^4/2 +x^5/5, from 0 to 3

Answer 13

well, at zero its 0; so at 3 ... then times it by sqrt(3)/36

Answer 14

Your method is confusing me a little amistre, but I think it is right
but I don't see what is wrong with sheena's setup, it is the same as I got

Answer 15

its confusing me too since we already got a height and im including it again ...

Answer 16

\[\frac{b}{h}=\frac{2}{\sqrt{3}}\]
\[b(h)=\frac{2}{\sqrt{3}}h\]
is how the base relates to height when I consider this :)

Answer 17

i am still geting 3.507 (is this right?)

Answer 18

I did it like\[y^2=(\frac y2)^2+h^2\implies h=\frac{\sqrt3}2y\]the area of each triangle is\[A(x)=\frac12bh=\frac{\sqrt3}4y^2=\frac{\sqrt3}4(3x-x^2)^2\]so the integral is\[\frac{\sqrt3}4\int_{0}^{3}(3x-x^2)^2\]which is what sheena had