Understanding Volume Calculations for Rotational Solids

In summary, RedBarchetta explains that the region bounded by the given curves is rotated about the x-axis. Then, y=-x+9x-18 is found. The volume of the solid is found by any method. When x is plugged into the equation for y, the mistake is made of thinking that y is a function of x. RedBarchetta solves the equation and shows us the steps he took to find the answer.
  • #1
RedBarchetta
50
1
1.Consider the given curves to do the following.
x=4+(y-3)[tex]^{2}[/tex], x=8

Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis. Sketch the region and a typical shell.

I'm lost on x being a function of y. How do you even enter these into a TI-83? Is there any way to make these easier?

Here's another one I've been working on that has caused me problems.
_________________________________________________________________
2.The region bounded by the given curves is rotated about the x-axis.
y=-x[tex]^{2}[/tex]+9x-18
y=0
Find the volume V of the resulting solid by any method.

First I graphed it:
htp://img377.imageshack.us/img377/3471/6338ol4.png (Just add another t in http)

cylindrical shell method
2 pi r h dr

I set up an integral. I get confused on determining the radius and height. If I'm rotating around the x-axis, I'm using y's. So, the radius should be y since it's centered around the y-axis. Then what is the shell height? The points at which the parabola crosses the curve are at x=3 and 6. So the shell height should be 6-y, but I think it should be where x = the equation.

But when I tried to single out x in the equation to get y as a function of x in y=-x[tex]^{2}[/tex]+9x-18, I couldn't calculate it.

Discs method
The area of one disc:
A(x)=[tex]\pi[/tex] * (-x[tex]^{2}[/tex]+9x-18)[tex]^{2}[/tex]

So the integral is
[tex]\pi[/tex] times the integral of (-x[tex]^{2}[/tex]+9x-18)[tex]^{2}[/tex] dx

Now the limits of integration should be from 3 to 6. I then integrated, plugged in the answer to my homework application which prompted me with a predictable "wrong" result. I've had no problems integrating, as I've completed 85% of my homework, but I have a hard time setting these problems up. Especially these odd-ball problems.

I'm sorry for my vague descriptions. It's hard to describe some of these things over the net. I appreciate any help. Thank you! :smile:
 
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  • #2
Hi RedBarchetta! :smile:
RedBarchetta said:
I'm lost on x being a function of y. How do you even enter these into a TI-83? Is there any way to make these easier?

You can always change any variable letter into any other letter … provided, of course, you remember to change them back at the end!

Just interchange x and y, to give y = 4+(x-3)², y=8. :smile:
Then what is the shell height? The points at which the parabola crosses the curve are at x=3 and 6. So the shell height should be 6-y, but I think it should be where x = the equation.

Forget the word "height" … you need the area of each shell, and the thickness of each shell.

Then you integrate over the thickness … sometimes it's height, sometimes it's radius, sometimes … :confused:

In this case, the thickness is not x, but dx.

Just think of it as "thickness", and you won't be confused! :smile:
But when I tried to single out x in the equation to get y as a function of x in y=-x+9x-18, I couldn't calculate it.

Sorry … I don't understand this … y is a function of x. :confused:

Show us your working on the integral, and then we can see where the mistake is. :smile:
 
  • #3
Thanks for the help Tiny-Tim! I figured both of these out now. :approve:
 

Related to Understanding Volume Calculations for Rotational Solids

1. What is the purpose of using calculus in volume problems?

Calculus allows us to accurately calculate the volume of irregular shapes and solids by breaking them down into smaller, more manageable pieces. This allows us to find the total volume of the object, even if it has a complex shape.

2. How is calculus used to find the volume of a solid?

To find the volume of a solid using calculus, we use the concept of integration. We break down the solid into infinitesimally small pieces, calculate the volume of each piece, and then sum up all the volumes to get the total volume of the solid.

3. Can calculus be used to find the volume of any shape?

Yes, calculus can be used to find the volume of any shape as long as we have a function that describes the shape. This function can be used to find the area of cross-sections of the solid, which can then be integrated to find the total volume.

4. How does calculus help in solving real-world volume problems?

Calculus is used in many real-world applications such as engineering, physics, and economics. It allows us to accurately find the volume of objects with irregular shapes, which is essential in designing structures, calculating fluid flow, and optimizing production processes.

5. Are there any practical examples of volume problems that require the use of calculus?

Yes, there are many practical examples of volume problems that require the use of calculus. For instance, calculating the volume of a water tank with a curved bottom, determining the volume of an irregularly shaped piece of land for construction purposes, or finding the volume of a chemical solution with varying concentrations at different depths.

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