Thread: Integration problem, Curve lenght, Area and Volume

1. Integration problem, Curve lenght, Area and Volume

Picture of ramp.

Problem statement: Curved edge of that ramp is y=(1/4)x3/2 Lenght of the ramp is 4m, width is 5m and height is 2m.
Calculate the lenght of the curve, then calculate the volume between the curve and floor and finally calculate center of mass in x direction.

Solution so far:
Step 1. Lenght of the curve is calculated using a formula deriving the function i get (3/8)x1/2 and then putting that into the formula i get a lenght of 4,512m.

Step 2. Volume is calculated using the slice method and my thinking is that the area of slice is acquired by integrating the function of the ramp with a definite integral between 4 and 0 and out of that i get an answer of 3,2m2
Volume is then a definite integral between 5 and 0 and integrating 3,2dy i get a volume of 16m3

Step 3. Center of mass Xc is calculated with My/A where My= so My is acquired by integrating x(1/4)x^3/2dx and out of that comes (1/14)x7/2 plugging that into the definite integral between 5 and 0 i get and answer of 19,96

Xc is then 19,96/3,2 which is 6,23m and that is just a weird answer. i would like to know where i got wrong
Thanks

2. Re: Integration problem, Curve lenght, Area and Volume

in step 3 the integral is from 0 to 4 instead of 0 to 5

$$\frac{\int _0^4\frac{1}{4}x x^{3/2}dx}{\int _0^4\frac{1}{4}x^{3/2}dx}$$

3. Re: Integration problem, Curve lenght, Area and Volume

the answer seems reasonable, but i would like to know why from 4 to 0 instead of 5 since the height of volume element is dy and in y-direction the lenght is 5m?

4. Re: Integration problem, Curve lenght, Area and Volume

I visualized this to be in the $\displaystyle xy$ plane.

in the $z$ direction we have $5$ units so the coordinates of the center of mass
would probably be something like

$\left\{\frac{20}{7},\frac{5}{8},\frac{5}{2}\right \}$

5. Re: Integration problem, Curve lenght, Area and Volume

ah, i see thank you very much