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

In summary, the conversation discusses finding the volume of a solid obtained by rotating a given region about a specific line using the washer method. The inner and outer radii are defined as the sum of the function and the distance from the line, respectively. The values of a and b are determined to be 0 and 1, and the final volume is evaluated to be $\pi(119/15)$.
  • #1
WDGSPN
6
0
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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  • #2
Hi WDGSPN (Wave), welcome to MHB!

In this problem, I would rotate it using the washer method. Are you familiar with it?
 
  • #3
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?
 
  • #4
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 $b$?
 
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  • #5
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.
 
  • #6
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)
 
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  • #7
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?
 
  • #8
WDGSPN said:
Would it be (1-(-4))^2?

No, just 4 - (-1) = 5.
 
  • #9
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! :)
 

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

1. What is the Integral Volume Problem?

The Integral Volume Problem is a mathematical concept that involves finding the volume of a three-dimensional shape using integration techniques. It is often used in calculus and physics to solve real-world problems.

2. How is the Integral Volume Problem solved?

The Integral Volume Problem is solved by setting up an integral that represents the volume of the shape. This integral is then evaluated using integration techniques such as the disk or shell method. The resulting value is the volume of the shape.

3. What types of shapes can be solved using the Integral Volume Problem?

The Integral Volume Problem can be used to solve for the volume of various types of shapes, such as cylinders, spheres, cones, and more complex shapes like toroids and frustums. It can also be used to find the volume of irregular shapes by breaking them down into smaller, simpler shapes.

4. What are some applications of the Integral Volume Problem?

The Integral Volume Problem has many practical applications, such as calculating the volume of liquids in containers, finding the mass of objects with varying densities, and determining the amount of material needed to construct a shape. It is also used in fields like engineering, architecture, and fluid mechanics.

5. What are some common mistakes to avoid when solving the Integral Volume Problem?

One common mistake when solving the Integral Volume Problem is forgetting to set up the integral correctly, resulting in an incorrect answer. It is also important to pay attention to the limits of integration and to use the correct integration method for the given shape. Additionally, it is crucial to double check the units of measurement to ensure the final answer is in the correct units.

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