# Thread: Solid of rev and Surface Area

1. ## 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.

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

2. ## Re: Solid of rev and Surface Area

What have you been able to do with this?

-Dan

3. ## 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

4. ## Re: Solid of rev and Surface Area

This is what I've got so far.

5. ## Re: Solid of rev and Surface Area

The axis of revolution is at the Y-axis.

6. ## 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 $\pi x^2 dy$. That's why you need to integrate $\pi\int_0^{15/2} x^2 dy$. 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?

7. ## Re: Solid of rev and Surface Area

Only the slant area, excluding the top and bottom

8. ## Re: Solid of rev and Surface Area

Okay, so you only need to do the integral I gave you.

9. ## Re: Solid of rev and Surface Area

I thought that formula is for volume? The surface area is 2 pi times all others?

10. ## 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, $\sqrt{dx^2+ dy^2}= \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$. Since the circumference of a circle, of radius r, is $2\pi r$, the surface area the product of those two lengths, $2\pi x\sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$. That's the formula you had before, right?

11. ## 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