Using Calculus to find Volume of 3-D Solids

In summary, the speaker has been using integration to find the volume of various objects such as cylinders. They stacked solid disks to a height of h and used the formula V=\pi*r^2*h to find the volume of a cylinder. However, when trying to find the volume of a sphere and cone using a similar method, they encountered difficulties. They realized that the radius of each disk does not decrease linearly and found the correct equations for finding the volume of a sphere and cone using integration.
  • #1
Weather Freak
40
0
Hey folks... I've been thinking about using integration to deal with the volumes of various objects such as:

Cylinders: For cylinders, I used integration to find the volume. I imagined the area of a circle, which is [tex]\pi*r^2[/tex] and then decided that if I stacked a series of solid disks (which have the area of a circle) to a height of h, then I would get the volume of a cylinder of height h:

[tex]\int_{0}^{h} \pi*r^2*dh[/tex]

And that yields the correct volume for a cylinder, [tex]V=\pi*r^2*h[/tex].

My question is, can I apply this method of thinking to find the area of a sphere and a cone?

Every time I try, I get the wrong answer.

Sphere:
I assume that I need to take a series of disks and add them up, as I did for the cylinder, but the cross sectional area will change as they get slightly smaller. So I tried this:

[tex]2 * \int_{0}^{r} \pi*r^2*dr[/tex]

My logic was that the radius will shrink to zero as you get to the outer most disk of the sphere, and then you have to double the answer to get the second half of the sphere. The problem is that this gives me [tex](2/3)\pi*r^2[/tex] and the real volume is 4/3, not 2/3.

And then comes the cone, and I'm not even sure where to start since I would think it would be similar to the sphere, but obviously it's not since they look so different.

So yes, I'm a bit confused, but if anyone can decipher what I've written and point me in the right direction, that would be greatly appreaciated. Thanks!
 
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  • #2
Sphere:
I assume that I need to take a series of disks and add them up, as I did for the cylinder, but the cross sectional area will change as they get slightly smaller. So I tried this:
[tex]2 * \int_{0}^{r} \pi*r^2*dr[/tex]



My logic was that the radius will shrink to zero as you get to the outer most disk of the sphere, and then you have to double the answer to get the second half of the sphere. The problem is that this gives me and the real volume is 4/3, not 2/3.

But the radius of each disk doesn't decrease linearly.
Since the equation of a sphere is x2+ y2+ z2= R2, if we let z range from -R to R, we can measure the radius of each disk along either x or y axis- let's take the x-axis so y= 0 and x2+ z2= 0 or [itex]x= \sqrt{R^2- z^2}[/itex].
The area of the disk is [itex]\pi \left(\sqrt{R^2-z^2}\right)^2[/itex]
The volume of the sphere is given by
[tex]\pi \int_{-R}^R R^2- z^2 dz[/tex].

Similarly, if a cone has height h and base radius R, we can look at the xz-plane. One side of the cone is the line running from (0,0,h) to (R, 0, 0). That has equation x= R- (R/h)z with y= 0. for any z, x is the radius of a disc with area [itex]\pi x^2= \pi(R- (R/h)z)^2= \piR^2(1- \frac{z}{h})^2[/tex]. The volume is given by
[tex]\pi R^2\int_0^h(1- \frac{z}{h})^2 dz[/tex]
 
  • #3
edit: nevermind, I found my mistake. i was confusing the decreasing radius of the disc with the radius of the sphere.
 
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  • #4
that makes sense, thank you!
 

1. How is calculus used to find the volume of 3-D solids?

Calculus is used to find the volume of 3-D solids by breaking down the shape into infinitely small pieces and calculating the volume of each piece. These small pieces are then added together using integration to find the total volume of the solid.

2. What is the formula for finding the volume of a 3-D solid using calculus?

The formula for finding the volume of a 3-D solid using calculus is V = ∫a^b A(x) dx, where A(x) represents the cross-sectional area of the solid at a particular point and a and b represent the limits of integration.

3. Can calculus be used to find the volume of any 3-D solid?

Yes, calculus can be used to find the volume of any 3-D solid as long as the shape can be broken down into infinitesimally small pieces. This includes common shapes such as cubes, spheres, and cylinders, as well as more complex shapes like cones and pyramids.

4. What are the benefits of using calculus to find the volume of 3-D solids?

Using calculus to find the volume of 3-D solids allows for more accurate calculations, as it takes into account the changing shape of the solid. It also allows for the calculation of volumes for more complex shapes that may not have a simple formula.

5. Are there any limitations to using calculus to find the volume of 3-D solids?

One limitation of using calculus to find the volume of 3-D solids is that it can be a time-consuming process, especially for more complex shapes. Additionally, the accuracy of the calculation depends on the accuracy of the integration and the precision of the measurements used for the shape.

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