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

[WDGSPN]

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

 

[Dethrone]

Hi WDGSPN (Wave), welcome to MHB!

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

 

[WDGSPN]

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?

 

[Dethrone]

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 $b$?

 

[WDGSPN]

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.

 

[Dethrone]

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)

 

[WDGSPN]

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?

 

[Prove It]

Would it be (1-(-4))^2?

No, just 4 - (-1) = 5.

 

[Dethrone]

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

 

[MarkFL]

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

 

[WDGSPN]

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)

 

[Dethrone]

Alright, so I got \pi(119/15)

Excellent! (Yes)

 

[WDGSPN]

Excellent! (Yes)

Thanks for the help! :)