Volume of solids of revolutions. (Integral problem)

[Mutaja]

Homework Statement

A function is given at y = $\sqrt{x-x^2}$

Find the volume of the solid of the revolution around the x axis. Not sure if this is the correct translation, but find the volume when you revolve the function around the x axis.

The object in question is a sphere. What does the radius of the sphere has to be?

The Attempt at a Solution

I am slightly familiar with the disc method in this case, but I am not familiar with the notes I've written from my lectures. This is what I've done so far.

The function goes from 0-1, and when revolved it forms a sphere.

Radius: $\sqrt{x-x^2}$
Area: ∏*$r^2$ = ∏($\sqrt{x-x^2}$)

V_d = ∏ $\sqrt{x-x^2}$ dx

V_t = ∫∏$\sqrt{x-x^2}$ dx (upper limit 1, lower limit 0).

To find the volume o this sphere, I have to solve this integration problem:

∫∏$\sqrt{x-x^2}$ dx (upper limit 1, lower limit 0)

Here's where I get a problem. I attempt to use the substitution method, and I get this equation:

u = x-$x^2$
∫$\sqrt{u}$ dx
∫$u^(1/2)$ dx

This equals $\frac{u^(3/2)}{\frac{3}{2}}$

And I substitute u for x and end up with the final equation:

$\frac{(x-x^2)^(3/2)}{\frac{3}{2}}$

When I set in limits for this equation and solve the problem, I get 0 - this is obviously wrong.

I'm fairly sure that I've integrated the problem wrong, but have I even attempted to integrate the correct expression?

Any help as always highly appreciated.

Edit: I have trouble with the formatting when I want to raise expressions to a higher power when I use latex.

 

[LCKurtz]

Homework Statement

A function is given at y = $\sqrt{x-x^2}$

Find the volume of the solid of the revolution around the x axis. Not sure if this is the correct translation, but find the volume when you revolve the function around the x axis.

The object in question is a sphere. What does the radius of the sphere has to be?

Complete the square and put it in standard form to see.

The Attempt at a Solution

I am slightly familiar with the disc method in this case, but I am not familiar with the notes I've written from my lectures. This is what I've done so far.

The function goes from 0-1, and when revolved it forms a sphere.

Radius: $\sqrt{x-x^2}$
Area: ∏*$r^2$ = ∏($\sqrt{x-x^2}$)

V_d = ∏ $\sqrt{x-x^2}$ dx

V_t = ∫∏$\sqrt{x-x^2}$ dx (upper limit 1, lower limit 0).

You forgot to square the radius.

Edit: I have trouble with the formatting when I want to raise expressions to a higher power when I use latex.

Put the exponent in braces {}.

 

[Mutaja]

Complete the square and put it in standard form to see.



You forgot to square the radius.



Put the exponent in braces {}.

Ah, ok, thanks. Will this be correct?

r = ($\sqrt{x-x^2}$)$^2$ = x-$x^2$

V_d = ∏ (x-$x^2$) dx

V_t = ∫∏ (x-$x^2$) dx

V_t = [∏ ($\frac{x^2}{2}$ - $\frac{x^3}{3}$ ] (upper limit 1, lower limit 0)

V_t = ∏ ($\frac{1^2}{2}$ - $\frac{1^3}{3}$)

= $\frac{∏}{6}$

Total volume of the sphere is $\frac{∏}{6}$.

The radius of the sphere is x-$x^2$.

The radius doesn't make sense to me, otherwise what I've done looks ok to me.

Thanks a lot for your help, I really appreciate it.

 

[LCKurtz]

Ah, ok, thanks. Will this be correct?

r = ($\sqrt{x-x^2}$)$^2$ = x-$x^2$

V_d = ∏ (x-$x^2$) dx

V_t = ∫∏ (x-$x^2$) dx

V_t = [∏ ($\frac{x^2}{2}$ - $\frac{x^3}{3}$ ] (upper limit 1, lower limit 0)

V_t = ∏ ($\frac{1^2}{2}$ - $\frac{1^3}{3}$)

= $\frac{∏}{6}$

Total volume of the sphere is $\frac{∏}{6}$.

The radius of the sphere is x-$x^2$.

The radius doesn't make sense to me, otherwise what I've done looks ok to me.

Thanks a lot for your help, I really appreciate it.

You have an equation $y = \sqrt{x-x^2}$. Square both sides and put it in the standard form for the equation of a sphere. That will give you its center and radius. The standard form for a sphere is$$
(x-x_0)^2 + (y-y_0)^2 = r^2$$Once you know the radius, you can check your answer for the volume using $V = \frac 4 3 \pi r^3$ and your radius.

 

[Mutaja]

You have an equation $y = \sqrt{x-x^2}$. Square both sides and put it in the standard form for the equation of a sphere. That will give you its center and radius. The standard form for a sphere is$$
(x-x_0)^2 + (y-y_0)^2 = r^2$$Once you know the radius, you can check your answer for the volume using $V = \frac 4 3 \pi r^3$ and your radius.

I'm sorry, but I'm lost.

Any chance we can take a quick recap?

$y^2$ = x - $x^2$

Put that into $$
(x-x_0)^2 + (y-y_0)^2 = r^2$$

And I get:

r = $\sqrt{x^2 +(x-x^2)}$

I'm confused as to which equations are important. Should I get y = or r =, should I insert for (x-$x^2$) for y etc...

I've been watching Khan Academy's videos on Disc method (rotating f(x) about x axis) and Volume of a sphere to try and get an understanding of the problem - and I do. The problem occur when I try to implement my own equation into his methods + try to get help from you.

I'm confusing myself, obviously, but please help me clear this up, and I'll do my best to understand it.

 

[haruspex]

$y^2$ = x - $x^2$

Put that into $$
(x-x_0)^2 + (y-y_0)^2 = r^2$$

And I get:

r = $\sqrt{x^2 +(x-x^2)}$

How do you get that? What happened to x₀ and y₀?
You are looking for a combination of constants, r, x₀ and y₀, which makes $(x-x_0)^2 + (y-y_0)^2 = r^2$ reduce to $y^2 = x - x^2$.

 

[Mutaja]

How do you get that? What happened to x₀ and y₀?
You are looking for a combination of constants, r, x₀ and y₀, which makes $(x-x_0)^2 + (y-y_0)^2 = r^2$ reduce to $y^2 = x - x^2$.

I just assumed x₀ and y₀ was 0.

 

[haruspex]

I just assumed x₀ and y₀ was 0.

Then don't.

 

[Mutaja]

Then don't.

I realize that, but it doesn't go well with what my notes says, nor what Khan Academy says. I find the formula in my book, but I don't understand how to use it as it's explained bad in my opinion. I also realize that it's very likely that it's well explained, I just can't get my head around it and understand it.

The formula is sound, but what is x0 and y0, and are they related to x and y? Like the general understanding I have of this is that x0 and y0 are starting coordinates and x and y are ending coordinates. But I can't put numbers into the equation from looking at the graph, because that would give x0 and y0 = 0, x = 1, y equals 1/2.

Any chance you can elaborate or explain it somehow? I'll do my absolute best to understand this.

 

[haruspex]

what is x0 and y0

$x^2 + y^2 = r^2$ is the equation of a sphere radius r centred at the origin. $(x-x_0)^2 + (y-y_0)^2 = r^2$ is also the equation of a sphere radius r. Where do you think its centre is?

 

[Mutaja]

$x^2 + y^2 = r^2$ is the equation of a sphere radius r centred at the origin. $(x-x_0)^2 + (y-y_0)^2 = r^2$ is also the equation of a sphere radius r. Where do you think its centre is?

I re did the whole problem, and wrote down the equations.

I ended up with just solving the equation for volume in respect to radius so:

V= $\frac{4}{3}$ ∏ $r^3$

equals: r = $\sqrt[3]{\frac{3v}{4∏}}$

Putting in my above answers for volume ($\frac{∏}{6}$) I get r = $\frac{1}{2}$.

That is exactly the same as what I get when I graph the function. It should be correct?

I still have no clue on how to use the x and y equation you're using. Is there any way you can explain it in a simpler way than what you've done so far?

 

[Mark44]

The curve that's being revolved around the x-axis is $y = \sqrt{x - x^2}$. If you square both sides of this equation, you get y² = x - x², or x² - x + y² = 0.

The graph of the first equation is the upper half of a circle. The graph of the second equation is the entire circle, not a sphere as others have said in this thread. When you rotate the first graph about the x-axis, you get a sphere.

Earlier in this thread LCKurtz recommended that you complete the square in this equation -- x² - x + y² = 0 -- to write it in standard form for a circle. When you do that, you'll see that the center is at (1/2, 0) and the radius is 1/2.

Since you didn't follow up on LCKurtz's suggestion, is it because you don't know how to complete the square? I would guess that this is one of the topics at Khan Academy. If you don't know this technique, it would be good for you to learn it.

 

[Mutaja]

Since you didn't follow up on LCKurtz's suggestion, is it because you don't know how to complete the square?

The short answer is yes.

To elaborate a little, I haven't got any formula or equation like that in my notes. The only place I can find it is in my book, which to my understanding explains it badly (as previously stated, that's most likely because of me, not the author of the book).

But the way I have solved the problem works, does it not? I will look into LCKurtz's suggestion later for sure, but right not I'll try to do more problems like this one to get a better understanding of how to think when solving them. Or maybe it would be better to start exploring his suggestion straight away? Because my next problem is sort of the same, only around the y-axis. I will probably post another thread shortly on that topic if I can't solve it myself.

Again, your help is much appreciated. Thanks a lot.

 

[haruspex]

The short answer is yes.

To elaborate a little, I haven't got any formula or equation like that in my notes. The only place I can find it is in my book, which to my understanding explains it badly (as previously stated, that's most likely because of me, not the author of the book).

But the way I have solved the problem works, does it not? I will look into LCKurtz's suggestion later for sure, but right not I'll try to do more problems like this one to get a better understanding of how to think when solving them. Or maybe it would be better to start exploring his suggestion straight away? Because my next problem is sort of the same, only around the y-axis. I will probably post another thread shortly on that topic if I can't solve it myself.

Again, your help is much appreciated. Thanks a lot.

'Completing the square' is a fundamental technique for dealing with quadratics. It's how one obtains the usual formula for solving them. I would recommend filling that gap in your learning ASAP.

 

[Mark44]

Or maybe it would be better to start exploring his suggestion straight away?

Yes, right away.