I have been exploring solids of revolutions. I am trying to find different ways of expressing them to calculate their areas and volumes. For example, if a revolve the function $y=x^{\frac{1}{2}}$ around the $x$-axis, it will form a solid. I have therefore added another dimension, or variable.
How would I define this solid in terms of a function of $x$ and $y$ that results in $z$?
This would allow to me integrate the multivalued function with iterated integrals.
Thank You
On
Thank you for your input. I already knew about the disc method for finding the volume of solids of revolution. You did do a good job of going in depth while explaining it to me. However I seek 3-dimensional methods, perhaps the ones you referred to at the end of your comment. I am writing an exploration about solids of revolution and need various ways of computing their volume (and surface area but that's for another time). I will try to explain to you my methods and hopefully you can further help me, but please excuse my formatting as I am relatively new to this website.
While learning about volumes in Cartesian coordinates. I recognized a pattern of finding the area, denoted dA, and then integrate the function by double integrals, where dA became dxdy. As such, it would theoretically become base (dA) times height(the function). And that is what the disc method is doing. Unfortunately, the majority of examples were volumes under a surface, which is not what I am trying to accomplish. I am trying to integrate my solid of revolution by finding its 3-dimensional equation.
For a function of the form: y=f(x), the corresponding equation that I came up with for rotating it around the x-axis is
$$f(x)^2=z^2+y^2$$ I used the circle equation because every point on the original function would become a circle on the yz-plane when revolved around the x-axis. You can correct me if this formula is wrong. From this point is elementary to solve for z and thus, have a function of the form f(x,y). I have graphed the function ln(x) revolved around the x-axis here.
https://www.desmos.com/calculator/sqa1agpk93
As you see, I plugged ln(x) into the equation and solved for z.
Now, my question becomes, what kind of integral do I need to integrate this function, hence find its volume.
I was thinking using the dA method to integrate the individual circle and then integrate over x? But that does seem very similar to the disc method, only that we are using the 3D Cartesian equation. Similarly we could use polar coordinates to find the area of individual circles. I was also thinking about using iterated integrals, double or triple (which do you think would work?), to integrate the solid with respect to x,y and z.
Thanks you very much for reading this and helping me, as I know the formatting is mediocre.
Ps: I had to write an "answer" because my post was too long for a comment. @UniS20
On
I think you want Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area A and the length of the path traced by its centroid R, i.e., 2πR. When composite areas are involved, the centroid is the weighted sum of the component centroids. The bottom line is that the volume is given simply by $V=2\pi RA$.
On
convert meridian
$$ y = \sqrt x$$
to an assemblage of meridians
$$ \sqrt{x^2+y^2} = r = f(z) $$
So to find corresponding volume from
$$ y = 2 + \sin x $$
change it to form
$$ r= 2 + \sin z $$
so that
$$ V = \pi\int _{z_1}^{z_2} r^2 dz = \pi \int_{r_1}^{r_2} r^2 dz $$
depending on whether you like to express $r$ in terms of $z$ or $z$ in terms of $r$ respectively.
I will solve your example using two variables and make some statements at the end about the general process.
You start with $y = x^{1/2}$ and you rotate about the $x$-axis. You do this in a way that the distance from the $x$-axis to your graph is constant. Let's call this distance (the height of your graph) $r$ s.t. $r=x^{1/2}$. We now have a solid that has been rotated into the $z$-plane, where every $yz$-plane contains the circular cross section of the solid.
The area of a circle is $\pi r^2$. This means the area of the circle that lies on the plane $x = 1$ is $\pi* {1^{1/2}}^2 =\pi$. By repeating this process for every value of $x$, we get the function $A=\pi {x^{1/2}}^2 = x\pi$ s.t. $A$ represents the area of the circle associated with each value of $x$.
By integrating $A$, you get the sum of the area under the curve (sum of areas of the circles), and hence the volume $V$: $$V=\int x\pi dx$$ Plugging in bounds will get you the volume of the solid between your choice of $x$ values.
For this method, in general, you will want to:
1.) Identify the radius of the circle, making note of the axis of revolution. For example, if you rotated $y = x^{1/2}$ along the $y$-axis, your radius would be $y^2$, obtained by arranging $y$ in terms of $x$.
2.) Write out Area as a function of $r$ and the axis of revolution (reducing the problem to two variables).
3.) Integrate along the axis of revolution to find volume.
I should note that this is only one method of finding the volume of solids of revolution, which is known as the disc method. There exists a slight modification of this method known as the washer method for hollow solids, and a shell method which sums the surface area of several cylinders. More complex methods, usually involving all three dimensions also exist.
By request, one such way is as follows:
Here is the method I think you want to know about
We understand that when we rotate a graph about the x-axis, we get circles on the $xz$-plane of radius $y=f(x)$. This graph of this circle can thus be denoted using $r^2=y^2+z^2$. But what does $r$ equal? In fact, $r$ here is still $y=x^{1/2}$ from the disc method, even though $y$ is now variable. This is because $y=x^{1/2}$ only when $z = 0$.
Now that we have our relations, we seek to find a order of integration to find the volume of the solid. I chose to again, first find the area of the circle and then integrate along the $x$-axis.
Let's consider how we would normally approach integrating a 2-dimensional function with height $y$ and length $x$. Normally, we opt to define $y$ in terms of $x$ and integrate the function $y=f(x)$ along the range of $x$ that we wish to find the area for. We can apply this same method to the $yz$ plane circle. We know $y^2+z^2={x^{1/2}}^2$. This is equivalent to $y=\pm \sqrt{x-z^2}$. By the nature of the graph, we end up with two functions, and hence two integrals, which we will look at later. For now, we consider what range of $z$ we wish to integrate these functions along. This range is the range for which the circle is defined, or the range where $\sqrt{x-z^2}$ is real. This is: $-r$ to $+r$, which is $-x^{1/2}$ to $+x^{1/2}$.
We now have the following 2 integrals: $$\int_{-x^{1/2}}^{x^{1/2}} \sqrt{x-z^2} dz$$ $$\int_{-x^{1/2}}^{x^{1/2}} -\sqrt{x-z^2} dz$$ Because the latter of the two is negative, the physical area of the circle is $$\int_{-x^{1/2}}^{x^{1/2}} \sqrt{x-z^2} dz - \int_{-x^{1/2}}^{x^{1/2}} -\sqrt{x-z^2} dz$$ which is equivalent to $$2\int_{-x^{1/2}}^{x^{1/2}} \sqrt{x-z^2} dz$$ We have now found the area of the circle.
Finally, we integrate this area along the range of $x$ for which we wish to find the volume of the solid of revolution. Moving the coefficient to the beginning of the expression, we get: $$V=2\int_{x_1}^{x_2} \int_{-x^{1/2}}^{x^{1/2}} \sqrt{x-z^2} dz dx$$ where $x_1$ and $x_2$ are real constants.
In this case, the $dA$ you speak of is $dz dx$. Our integral can be generalized as $$\iint_R f(x,z) dA$$ In fact, by rearranging the order of integration, you can change the variables that $dA$ represents. This is because the most general formula of volume is $$V=\iiint_S dV$$ Because the first integral is the integral of $1$, this integral is usually left out. Theory-wise, however, this form is nice because it shows you that different equations for volume of the solid are really just rearranged orders of integration of this equation (often accompanied by simplifications done through geometrical observation or other tricks, such as Jacobian coordinate changes, but this is another topic).
Apologies for misinterpreting your question the first time, and I hope this helped!
EDIT: Added another method as requested by OP