Find the volume of the described solid S
- Thread starter flakine
- Start date
Good call, pka. I was thinking along those lines, but I thought it was me.
If you figure out what the problem is supposed to be, try using the area
of an equilateral triangle formula. \(\displaystyle \L\\\frac{\sqrt{3}}{4}y^{2}\)
Let your function be the base side of the triangle.
I assume they mean the parabola \(\displaystyle y=9x^{2}\)
May I ask what is the name and author of this text?.
If you figure out what the problem is supposed to be, try using the area
of an equilateral triangle formula. \(\displaystyle \L\\\frac{\sqrt{3}}{4}y^{2}\)
Let your function be the base side of the triangle.
I assume they mean the parabola \(\displaystyle y=9x^{2}\)
May I ask what is the name and author of this text?.
Do what they suggest.
Solve your equation for x: \(\displaystyle x=\frac{sqrt{y}}{3}\)
Let x be the base of your triangle and insert it in your area formula:
\(\displaystyle \frac{\sqrt{3}}{4}x^{2}\)
\(\displaystyle \L\\4\cdot\frac{\sqrt{3}}{4}\int_{0}^{81}\frac{y}{9}dy\)
EDIT: typo.
Solve your equation for x: \(\displaystyle x=\frac{sqrt{y}}{3}\)
Let x be the base of your triangle and insert it in your area formula:
\(\displaystyle \frac{\sqrt{3}}{4}x^{2}\)
\(\displaystyle \L\\4\cdot\frac{\sqrt{3}}{4}\int_{0}^{81}\frac{y}{9}dy\)
EDIT: typo.
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,216
First of all, your notation ... could it actually be 9x<sup>2</sup> < y < 81 ???
this would describe the region above (and on) the parabola y = 9x<sup>2</sup>, and below (and on) the line y = 81.
if this is the case ...
cross sections perpendicular to the y-axis are equilateral triangles ... each side of an equilateral triangle has length = 2x
using the equation provided by galactus ... \(\displaystyle \L A = \frac{\sqrt{3}}{4}(2x)^2 = \sqrt{3} x^2\)
since y = 9x<sup>2</sup>, x<sup>2</sup> = y/9
the volume of a representative slice is \(\displaystyle dV = \frac{\sqrt{3}}{9} y dy\)
so ... \(\displaystyle \L V = \int_0^{81} \frac{\sqrt{3}}{9} y dy = \frac{729\sqrt{3}}{2}\)
this would describe the region above (and on) the parabola y = 9x<sup>2</sup>, and below (and on) the line y = 81.
if this is the case ...
cross sections perpendicular to the y-axis are equilateral triangles ... each side of an equilateral triangle has length = 2x
using the equation provided by galactus ... \(\displaystyle \L A = \frac{\sqrt{3}}{4}(2x)^2 = \sqrt{3} x^2\)
since y = 9x<sup>2</sup>, x<sup>2</sup> = y/9
the volume of a representative slice is \(\displaystyle dV = \frac{\sqrt{3}}{9} y dy\)
so ... \(\displaystyle \L V = \int_0^{81} \frac{\sqrt{3}}{9} y dy = \frac{729\sqrt{3}}{2}\)