Volume of revolution where am i going wrong?

In summary, the conversation discusses finding the volume of a solid obtained by rotating a region bounded by y=x^2, y=4, x=0 about the line x=-2 using the shell method. The person initially gets the correct answer using the disk method, but is stuck on the shell method. They ask for help and share their work, but are unsure what part of the integral is wrong. Eventually, they realize there should be a hole in the center of the volume and discuss whether or not the shell method accounts for it. They also mention that Khan Academy may have the wrong answer and discuss potential mistakes in their calculations.
  • #1
helpmeplz!
27
1
Hey guys I'm stuck on this problem. Its an easy one but I need some help..

It's asking for the volume of the solid obtained by rotating the region bounded by y=x^2, y=4, x=0 about the line x= -2 using the shell method.

I got the answer correct using the disk method (answer is 136 pi/3)..

With shell method I get integral 2pi (2+x) (4-x^2) dx from x=0 to x=2 which is 88pi/3. Can someone please help me out of this mess?
 
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  • #2
You need to supply your work details.
 
  • #3
What do you mean? My question is what part of the integral that I set up is wrong and why? Not the evaluating of the integral which I'm using a calculator for anyway.
 
  • #4
jbrussell93 said:
Hmmm... are you sure that is the correct answer? I got ##\frac{88 \pi}{3}## for both

Using slices, I get integral pi * (2+sqrty)^2 dy from y=0 to 4. which gives 136 pi/3?
 
  • #5
There should be a hole in the center due to the gap from x=-2 to x=0. I'll leave the rest up to you
 
  • #6
jbrussell93 said:
There should be a hole in the center due to the gap from x=-2 to x=0. I'll leave the rest up to you

Right, but the shell method takes that into account doesn't it? My integral only foes from 0 to 2..
 
  • #7
Yes the shell method does, so I think you got the correct answer there. Think about the disk method and what your volume would look like. (Yours is missing the hole)
 
  • #9
Yeah, it looks like it... Unless I'm wrong, which is entirely possible :P
 
  • #10
Actually, I don't think he ever specifies x=0 as a bound. So assuming x=-2 is the bound instead, it would be correct.
 
  • #11
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Thread moved from Mathematics > Calculus to
Homework & Coursework Questions > Calculus & Beyond Homework

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FAQ: Volume of revolution where am i going wrong?

What is the volume of revolution?

The volume of revolution is the volume obtained by rotating a two-dimensional region about a line in three-dimensional space. It is also known as the solid of revolution.

How is the volume of revolution calculated?

The volume of revolution is calculated using the formula V = π∫abf(x)^2dx, where a and b are the limits of integration and f(x) is the function used to generate the solid of revolution.

What are some common methods for finding the volume of revolution?

Some common methods for finding the volume of revolution include the disk method, the shell method, and the washer method. These methods involve slicing the solid of revolution into thin disks or shells and then using calculus to find the volume of each slice and adding them together.

What are some common mistakes when calculating the volume of revolution?

Some common mistakes when calculating the volume of revolution include using the wrong limits of integration, forgetting to square the function, and using the wrong method for the given solid of revolution. It is important to carefully set up the integral and double check all calculations to avoid these mistakes.

Are there any real-world applications for the volume of revolution?

Yes, the volume of revolution has many real-world applications in fields such as engineering, physics, and architecture. For example, it can be used to calculate the volume of objects such as pipes, bottles, and tunnels, which are often modeled as solids of revolution.

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