Negative derivative instead of positive

[Karol]

Homework Statement

Homework Equations

Differential of a product:
$$d(uv)=u\cdot dv+v\cdot du$$

The Attempt at a Solution

$$dV=\pi \left[ -\frac{1}{x}x^2+2x\left(-\frac{x}{3} \right) \right]dx=-\pi x^2dx$$
If dx>0 dV<0, it's wrong, the volume increases

 

[fresh_42]

Homework Statement

View attachment 209459

Homework Equations

Differential of a product:
$$d(uv)=u\cdot dv+v\cdot du$$

The Attempt at a Solution

$$dV=\pi \left[ -\frac{1}{x}x^2+2x\left(-\frac{x}{3} \right) \right]dx=-\pi x^2dx$$
If dx>0 dV<0, it's wrong, the volume increases

I think you've made a mistake in your differentiation. And then, what are the relevant values of depth? Can there be more than 10 inches?

 

[Karol]

$$V=\pi\left( 10x^2-\frac{1}{3}x^3 \right)~\rightarrow~V'=\pi x (20-x)$$
When, for example, x=8 ⇒ V=469⅓π, x=2 ⇒ V=37⅓π
So x is measured from the bottom and it can't pass 10
But now the derivative is positive and i think it's fine.

 

[fresh_42]

$$V=\pi\left( 10x^2-\frac{1}{3}x^3 \right)~\rightarrow~V'=\pi x (20-x)$$
When, for example, x=8 ⇒ V=469⅓π, x=2 ⇒ V=37⅓π
So x is measured from the bottom and it can't pass 10
But now the derivative is positive and i think it's fine.

Yes, it's a (bottom opened) parabola. This makes sense, as it is the formula of the volume of a sphere, and if you could go above the equator, the volume will decrease again compared to the height or depth.

 

[Mark44]

This makes sense, as it is the formula of the volume of a sphere, and if you could go above the equator, the volume will decrease again compared to the height or depth.

Actually, the volume will continue to increase (not decrease) as the container is filled, but will do so at a decreasing rate. I think I understand what you meant, but the words you chose didn't match the situation.

 

[Karol]

The volume of a sphere is $~V=\frac{4}{3}\pi r^3$ and we are given $~V=\pi\left( 10x^2-\frac{1}{3}x^3 \right)$
Is the term 10x² and the ⅓x³ instead of 4/3x³ because the bowl is under the equator?
How do i know to calculate that?

 

[fresh_42]

The volume of a sphere is $~V=\frac{4}{3}\pi r^3$ and we are given $~V=\pi\left( 10x^2-\frac{1}{3}x^3 \right)$
Is the term 10x² and the ⅓x³ instead of 4/3x³ because the bowl is under the equator?
How do i know to calculate that?

The volume of what's in the bowl is $\displaystyle V = \frac{\pi\, x^2}{3}(3r-x)$

 

[Karol]

How to translate from one to another?
The formula in Wikipedia $~V=\frac{\pi h}{6}(3a^2+h^2)=\frac{\pi x}{6}(3a^2+x^2)$
$$a^2+(r-x)^2=r^2~\rightarrow~r=\frac{a^2+x^2}{2x}$$
How did you get to the formula:
$$\displaystyle V = \frac{\pi\, x^2}{3}(3r-x)$$

 

[fresh_42]

I didn't bother a lot about the proof of this formula. I simply changed languages on Wiki and took the one that fitted best. As we're in the advanced section here, you could try and solve the corresponding integral. It can also be found on this other page, but I'm too lazy to copy and translate it (and risk typos). I'm sure if you google "spherical segment" you will find plenty of proofs.

 

[Karol]

you could try and solve the corresponding integral. It can also be found on this other page,

Which page, in this site?
Where is the integral?
Thank you fresh_42

 

[fresh_42]

Which page, in this site?
Where is the integral?

With Pythagoras you can find the equation for the circle segment above the x-axis for the shape of the cap, then take the integral for the volume of rotational bodies and you'll get the formula. You could switch languages by yourself, depending on which language you understand. But it's math, so it isn't really needed to understand everything. However, it would be a good and not too difficult exercise to derive the formula for practice.