Help interpreting a question: is this asking for the volume of a sphere?

[KrabLord]

So, I think I’m supposed to find the volume of a sphere by taking square slices. Thus, where r is radius, the function which yields the area in terms of x is:

A(x) = diameter^2 = (2 • sqrt(r^2 - x^2))^2 .

Is this line of reasoning correct?

The given answer is:

So I know that I’ve arrived at the right formula, but I am curious as to if my line of reasoning which lead me there is correct.

 

[MarkFL]

The volume in question isn't spherical. I would begin with:

[MATH]dV=(2y)^2\,dx=4y^2\,dx=4(a^2-x^2)\,dx[/MATH]
The base of each square slice extends from \(-y\) to \(y\) for a length of \(2y\), and since the slices are square, their area is \((2y)^2\), and then we need to express \(y^2\) as a function of \(x\) knowing:

[MATH]x^2+y^2=a^2\implies y^2=a^2-x^2[/MATH]
Hence:

[MATH]V=4\int_{-a}^{a} a^2-x^2\,dx=8\int_0^a a^2-x^2\,dx=8\left[a^2x-\frac{1}{3}x^3\right]_0^a=8\left(\frac{2}{3}a^3\right)=\frac{16}{3}a^3[/MATH]

 

[KrabLord]

The volume in question isn't spherical. I would begin with:

[MATH]dV=(2y)^2\,dx=4y^2\,dx=4(a^2-x^2)\,dx[/MATH]
The base of each square slice extends from \(-y\) to \(y\) for a length of \(2y\), and since the slices are square, their area is \((2y)^2\), and then we need to express \(y^2\) as a function of \(x\) knowing:

[MATH]x^2+y^2=a^2\implies y^2=a^2-x^2[/MATH]
Hence:

[MATH]V=4\int_{-a}^{a} a^2-x^2\,dx=8\int_0^a a^2-x^2\,dx=8\left[a^2x-\frac{1}{3}x^3\right]_0^a=8\left(\frac{2}{3}a^3\right)=\frac{16}{3}a^3[/MATH]

MarkFL, you are my hero. Thank you so much.

 

[Steven G]

In the end possibly you are correct that you are finding the volume of a sphere. However you just can't assume it is a sphere, guess it is a sphere or do anything but think hard about what it will look. Can you picture the volume? Is it a sphere?

In this problem you can find the volume even with out knowing exactly what the 3d figure will look like. You know, or should know, that the left most slice will be a very very small square, then a little bigger square,...eventually a square with sides equal to the radius. In fact you can stop here and double this volume as the volume to the right of that biggest square will just be the mirror image of what you just found.

Stop trying to do these problem without thinking about what is going on.

 

[KrabLord]

In the end possibly you are correct that you are finding the volume of a sphere. However you just can't assume it is a sphere, guess it is a sphere or do anything but think hard about what it will look. Can you picture the volume? Is it a sphere?

In this problem you can find the volume even with out knowing exactly what the 3d figure will look like. You know, or should know, that the left most slice will be a very very small square, then a little bigger square,...eventually a square with sides equal to the radius. In fact you can stop here and double this volume as the volume to the right of that biggest square will just be the mirror image of what you just found.

Stop trying to do these problem without thinking about what is going on.

I’m studying Gilbert strang’s calculus text in my free time, not for a class. I am thinking about what is going on. I happened to fail to land on the wrong visualization, but that’s by virtue of incompetence, not an absence of effort. In this case, I was visualizing cubes, not squares, which clearly indicates a lack of rigor in my thinking. I will claim incompetence, but I am trying.

 

[Steven G]

Remember that dx is small, so you can't have cubes.

 

[MarkFL]

Here's a video that will help you visualize the shape of the solid being discussed:

 

[KrabLord]

Here's a video that will help you visualize the shape of the solid being discussed:

Very nice!