# Solid of rev and Surface Area

• Apr 9th 2013, 07:00 PM
ggallego
Solid of rev and Surface Area
I think I posted it in a wrong forum, being a newbie here and needs some help on my math problem. Thanks for all the help guys.

http://i45.tinypic.com/339r6s6.jpg

Hi I'm stuck with my homework for solid of revolution and surface area of this figure Trapezoid with slope of -5/L. R2 = 2L and R1 = 3L/2. ANy help is appreciated.

/ ----- |---------\
/ -----| ---------\ slope = -5/L
/_____|____|__\
...............3L/2 ...2L

a. Find the equation in terms of L (vol and surface area)
b. Vol as function of L
c. Value of L that minimizes the volume
d. Minimum Volume
• Apr 10th 2013, 06:34 AM
topsquark
Re: Solid of rev and Surface Area
What have you been able to do with this?

-Dan
• Apr 10th 2013, 06:45 AM
hollywood
Re: Solid of rev and Surface Area
What is the axis of rotation? Isn't part (b) repeating the first half of part (a)?

And doesn't the volume decrease as L decreases, so (c) and (d) have no solution, assuming L>0?

- Hollywood
• Apr 10th 2013, 07:27 AM
ggallego
Re: Solid of rev and Surface Area
This is what I've got so far.

http://i47.tinypic.com/wld3rm.jpg
• Apr 10th 2013, 07:28 AM
ggallego
Re: Solid of rev and Surface Area
The axis of revolution is at the Y-axis.
• Apr 10th 2013, 07:52 AM
HallsofIvy
Re: Solid of rev and Surface Area
Okay, taking the origin at the center of the base, the side is a line with slope -5/L passing through (2L, 0). That has equation y= (-5/L)(x- 2L). When x= 3L/2, y= (-5/L)(3L/2- 2L)= (-5/L)(-3L/2)= 15/2. Rotating around the y- axis, you have disks with radius x, thickness dy, so area $\displaystyle \pi x^2 dy$. That's why you need to integrate $\displaystyle \pi\int_0^{15/2} x^2 dy$. If y= (-5/L)(x- 2L), what is x as a function of y? Now, when you say "surface area" do you mean the slant area only or do you want to incude the area of the top and bottom?
• Apr 10th 2013, 08:19 AM
ggallego
Re: Solid of rev and Surface Area
Only the slant area, excluding the top and bottom
• Apr 10th 2013, 08:36 AM
HallsofIvy
Re: Solid of rev and Surface Area
Okay, so you only need to do the integral I gave you.
• Apr 10th 2013, 08:44 AM
ggallego
Re: Solid of rev and Surface Area
I thought that formula is for volume? The surface area is 2 pi times all others?
• Apr 10th 2013, 12:32 PM
HallsofIvy
Re: Solid of rev and Surface Area
Oh, blast! That's right. Then you can do this: a line with slope m is of the form y= mx+ b. If we increase x by dx then we increase (decrease for negative m) y by y+ dy= m(x+ dx)+ b subtracting y from the left side and mx+ b from the right we get dy= m dx (which is just saying, again, that for a straight line the slope is the same as the derivative). A right triangle with legs dx and dy has hypotenus of length, by the Pythagorean theorem, $\displaystyle \sqrt{dx^2+ dy^2}= \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$. Since the circumference of a circle, of radius r, is $\displaystyle 2\pi r$, the surface area the product of those two lengths, $\displaystyle 2\pi x\sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$. That's the formula you had before, right?
• Apr 10th 2013, 12:46 PM
ggallego
Re: Solid of rev and Surface Area
Yep, but using all the equation and applying my limits, I was getting negative results.

2 Pi X ( 1 + (dy/dx)^2)^1/2 dx from 0 to 5/2 for Y = 5X + 10