1. ## Rotational Vol Question.

I´ll try to explain this problem. Trying to solve this for the last 40minutes now.

Equation: y = x^(0,5)

The area under this equation starts to rotate around the X-axis and gives you the rotational Volume (B).

When calculating the above mentioned volume: Int: from (a = 0) to (b), There is a line from the point b parallell to the X-axis ( called line d), that line crosses the y-axis. The area between: Y-axis; This line (d), and the area "above" the y = x^0,5 graph starts to rotate around the y axis. This gives you volume (A).

Find x when these areas have the same volumes ( A = B).

The equation for calculating volume rotating around X-axis is:
INT: (pi)*(y^2):dx from a to b (on the x-axis as limes a = 0)

The equation for calculating volume rotating around y-axis is: Int:2*pi*xy:dx from a to b (on the X-axis ALSOOO).

At first I thought, never mind the rotation, just solve so the areas a equal, but that was not very clever, points further from the axises give larger values for volume then closer points do.

I keep getting the answer x = 0,625^2 but that is wrong, THE CORRECT SOLUTION IS (6.25;2.5)
Hope I discribed this problem understandably.

2. It's a little difficult following what you mean - partly because you talk about the "area" of A whereas I think you mean "volume," and also because I don't know what the "a" value is supposed to signify. Is "a" = 0? And when you ask for the value of x, do you mean the value of b to make the two volumes equal?

I will assume that the problem is to find a value for b such that the volume B (which is the defined by rotating y = sqrt(x) about the x axis, and finding the volume from x = 0 to x = b) is equal to the volume A (defined as rotating y = sqtr(x) about the y axis, and finding the volume from y = 0 to y = sqrt(b). If I have that right, then the answer for b turns out to be 6.25. I can help you thorugh this, but before I do please verify that I have interpreted the problem correctly.

3. ## You are right

Yes, you are correct, my mistake talking about the areas when the question was about volume. Please How did you get the answer?

4. Given the function y = sqrt(x): the volume of B is:

$\displaystyle B = \int _ 0 ^b y^2 dx = \int _0 ^b x dx = \frac {b^2} 2$

The volume of A is:
$\displaystyle A = \int _0 ^{\sqrt b} x^2 dy = \int _0 ^{\sqrt b} y^4 dy = \frac 1 5 (\sqrt b) ^5 = \frac 1 5 b^{ \frac 5 2}$

Set volume A = volume B, and solve for the value of b.

5. ## T

That was nice and simple