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[SOLVED] Petal area problem

DeusAbscondus

Active member
Jun 30, 2012
176
I've posted an attachment with the problem set forth,
but I can't seem to get the next part loaded within my set upload limit.

Any help would be appreciated.

D'Abs
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I think you need to write the volume V in cm3 as:

$\displaystyle V=15\cdot55^2-4\cdot100\cdot15\int_0^{\frac{1}{4}}\frac{\sqrt{x}}{2}-4x^2\,dx$

Do you see why?
 

DeusAbscondus

Active member
Jun 30, 2012
176
I think you need to write the volume V in cm3 as:

$\displaystyle V=15\cdot55^2-4\cdot100\cdot15\int_0^{\frac{1}{4}}\frac{\sqrt{x}}{2}-4x^2\,dx$

Do you see why?
But would that not be equivalent to multiplying the entire volume of the block by the volume of the cut out 3d flower?

$$1. 15\cdot 55^2 \text {is the volume of the uncut prism and the integrand is the curve }y=\frac{\sqrt{x}}{2}-y=4x^2$$

I will have to take the current screenshot down and put the next one up; awkward, but space won't allow for both at the same time.

thanks heaps for helping,
D'abs
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I am taking the volume of the prism and subtracting from it the volume of the leaf design which is cut through it.
 

DeusAbscondus

Active member
Jun 30, 2012
176
I am taking the volume of the prism and subtracting from it the volume of the leaf design which is cut through it.

Ah, that is my problem: I can 't see why one may not derive the volume of design without reference to the volume of the block from which it is "hewn", except that datum of depth: in this case 15cm.
The references to width and length, to the volume of entire block, seem to be entirely beside the point:
surely all you need is:
- the area of one leaf;
- knowledge that each leaf is identical AND that the front and back of the design are identical; and finally,
- depth of block

then it is simply:
$$4\cdot 15cm \int^{1/4}_0 [f(x)-g(x)] dx $$

Isn't it?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I need to amend my previous statement to say

$\displaystyle V=15\cdot55^2-4\cdot100^2\cdot15\int_0^{\frac{1}{4}}\frac{\sqrt{x}}{2}-4x^2\,dx$

I get 32,875 cm3.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Ah, that is my problem: I can 't see why one may not derive the volume of design without reference to the volume of the block from which it is "hewn".
You could, but it would be much more difficult I think.
 

DeusAbscondus

Active member
Jun 30, 2012
176
You could, but it would be much more difficult I think.
My notes say:

"The volume of concrete in the block:
= face area x depth
= 0.03288
Therefore, the concrete required is about 0.03288 m^3 or 33000cm^3"

This is what i mean: "face area x depth" should mean, on the face of it, the result one obtains from integrating the difference between the given functions by the depth ie: 15.

Bringing in the volume of the entire square just confuses me.
Still confused.
Thanks anyway,
D'abs
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If you take the area of the leaf, multiply it by the depth of the block, you have the volume removed from the block. This is why I took the volume of the block, then subtracted that volume which was removed to get the remaining volume.