Volume of Solid Rotated Around y=-4 for y=x^7, y=1, and y-axis

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    Integral Volume
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Discussion Overview

The discussion revolves around finding the volume of a solid obtained by rotating the region bounded by the curves y=x^7, y=1, and the y-axis about the line y=-4. The conversation includes the application of the washer method and the calculation of inner and outer radii for the volume integral.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant asks for help in calculating the volume of the solid of revolution using the washer method.
  • Another participant suggests using the washer method and asks if the original poster is familiar with it.
  • There is a discussion about identifying the inner and outer radii for the volume calculation, with one participant proposing the inner radius as ((x^7)+4)^2.
  • Clarification is provided that the inner radius is actually (x^7 - (-4))^2, emphasizing the geometric interpretation of the distance from the curve to the line y=-4.
  • Participants discuss the outer radius, with one suggesting it is (1-(-4))^2, while another corrects this to simply 5.
  • The final volume integral is presented as V=π∫(5)^2-(x^7+4)^2 dx, leading to a calculated volume of π(119/15).

Areas of Agreement / Disagreement

Participants generally agree on the method of using the washer method and the setup of the volume integral, but there are some uncertainties regarding the correct expressions for the inner and outer radii, particularly in the early stages of the discussion.

Contextual Notes

Some participants express uncertainty about the definitions of inner and outer radii, and there are unresolved aspects regarding the simplification of the outer radius and the final evaluation of the integral.

WDGSPN
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Help with the following problem will be much appreciated:

Find the volume of the solid obtained by rotating the region bounded by y=x^7, y=1, and the y-axis about the line y=−4
 
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Hi WDGSPN (Wave), welcome to MHB!

In this problem, I would rotate it using the washer method. Are you familiar with it?
 
Rido12 said:
Hi WDGSPN (Wave), welcome to MHB!

In this problem, I would rotate it using the washer method. Are you familiar with it?

Hi Rido12 , I am not quite familiar with it, can you explain how you would use it please?
 
If you rotate the region you are given about $y=-4$, you are producing washers and you can find the volume by:

$$V=\pi \int_a^b (\text{outer radius})^2-(\text{inner radius})^2 \,dx$$

What is the inner and outer radius, and what is $a$ and $b$?
 
Last edited:
Rido12 said:
If you rotate the region you are given about $y=-4$, you are producing washers and you can find the volume by:

$$V=\pi \int_a^b (\text{outer radius})^2-(\text{inner radius})^2 \,dx$$

What is the inner and outer radius, and what is $a$ and $b$?

Uhmmm would b=1 and a=0 ? I was thinking the inner radius would be ((x^7)+4)^2 but I'm not sure.
 
Correct on both. The inner radius is $(x^7-(-4))$ because the radius is the sum of both the function $x^7$ and the distance the line $y=-4$ from the x-axis, geometrically. So, $(\text{inner radius})^2=(x^7-(-4))^2$

Now, what is the outer radius? (Wondering)
 
Last edited:
Rido12 said:
Correct on both. The inner radius is $(x^7-(-4))^2$ because the radius is the sum of both the function $x^7$ and the distance the line $y=-4$ from the x-axis, geometrically.

Now, what is the outer radius? (Wondering)

Would it be (1-(-4))^2?
 
WDGSPN said:
Would it be (1-(-4))^2?

No, just 4 - (-1) = 5.
 
WDGSPN said:
Would it be (1-(-4))^2?

Right, that is $(\text{outer radius})^2$, so now you have

$$V=\pi \int_0^1(5)^2-(x^7+4)^2\,dx$$

What does that evaluate to? :D
 
  • #10
Prove It said:
No, just 4 - (-1) = 5.

Rido12 and the OP have been working with the square of the inner and outer radii, and while the actual outer radius in this case is just 5, the OP was giving the square of this radius, which was given correctly, although not yet simplified. :D
 
  • #11
Rido12 said:
Right, that is $(\text{outer radius})^2$, so now you have

$$V=\pi \int_0^1(5)^2-(x^7+4)^2\,dx$$

What does that evaluate to? :D

Alright, so I got \pi(119/15)
 
  • #12
WDGSPN said:
Alright, so I got \pi(119/15)

Excellent! (Yes)
 
  • #13
Rido12 said:
Excellent! (Yes)
Thanks for the help! :)
 

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