What have you been able to do with this?
-Dan
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
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 . That's why you need to integrate . 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?
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, . Since the circumference of a circle, of radius r, is , the surface area the product of those two lengths, . That's the formula you had before, right?