Calculating Volume by Integration

In summary, the conversation is discussing how to find the volume of a solid obtained by rotating the region bounded by the curves x = 2 sqrt(y), x = 0, y = 9 about the y-axis. The setup involves integrating with respect to y, with the outer radius being x = 2 sqrt(y) and the inner radius being x = 0. The final answer is found to be 162π, after a mistake is corrected in the integration process.
  • #1
johnhuntsman
76
0
Find the volume of the solid obtained by rotating the region bounded by the curves:

x = 2 sqrt(y), x = 0, y = 9; about the y-axis

I have this graphed and everything so I'm not sure there's a need to worry about that.

Setup (integrating with respect to y):
Outer radius = x = 2 sqrt(y). Inner radius = x = 0. Bounded by y = 9 and 0.

9
∫π[2 sqrt(y)]^2 dy
0

Which is equivalent to:
_______9
(2πy^2)| = 36π
_______0

However, my book says the answer is 162π. What am I doing wrong?
 
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  • #2
Ahhh now you might've missed this :

x = 2[itex]\sqrt{y}[/itex]

So your volume integral becomes :

[itex]\pi \int_{0}^{9} 4y \space dy[/itex]

Since [itex]f^2(y) = 4y[/itex]
 
Last edited:
  • #3
Zondrina said:
Ahhh now you might've missed this :

x = 2[itex]\sqrt{y}[/itex]

So your volume integral becomes :

[itex]\int_{0}^{9} 4y \space dy[/itex]

Since [itex]f^2(y) = 4y[/itex]

That doesn't change what I get at the end. I'll still factor out the 4 and the π and what's left in the integral will still come to be (y^2) / 2. Multiply that by what's outside the integral and I still have 2πy^2.
 
  • #4
johnhuntsman said:
That doesn't change what I get at the end. I'll still factor out the 4 and the π and what's left in the integral will still come to be (y^2) / 2. Multiply that by what's outside the integral and I still have 2πy^2.

Oh whoops, in my last post i meant your integral is really :

[itex]π \int_{0}^{9} 4y \space dy[/itex]

So, integrating it you get :

[itex]π \int_{0}^{9} 4y \space dy = 4π \int_{0}^{9} y \space dy = 2π[y^2]_{0}^{9} \space dy[/itex]

Do you see it now?
 
  • #5
Aah. I know what I did. I just thought that came out to 36π for some stupid reason. Alright thanks. Sorry about that. I'm a bit of a dope : D
 

What is "Volume by Integration"?

"Volume by Integration" is a mathematical method used to find the volume of an irregularly shaped object or space by breaking it down into infinitely small pieces and summing them up using calculus.

How is "Volume by Integration" different from other methods of finding volume?

Unlike other methods such as using geometric formulas or physical measurements, "Volume by Integration" can be used to find the volume of any irregularly shaped object or space, even if its shape is constantly changing.

What are the steps involved in using "Volume by Integration" to find volume?

The first step is to define the object or space in terms of a mathematical function. Then, using calculus, you can find the area of cross-sections of the object or space at different points. Finally, you integrate these areas over the entire length or height of the object or space to find the total volume.

What are the applications of "Volume by Integration" in science?

"Volume by Integration" is commonly used in physics, engineering, and other sciences to find the volume of irregularly shaped objects or spaces. It is also used in fluid mechanics to calculate the volume of liquid or gas in a container with varying shapes or sizes.

Are there any limitations to using "Volume by Integration"?

While "Volume by Integration" is a powerful method, it can be challenging and time-consuming for complex shapes and functions. It also requires a good understanding of calculus and mathematical functions. Additionally, it may not be suitable for objects or spaces with constantly changing shapes or sizes.

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