Single Var Calculus - Volumes of Revolution

[anonymity]

Single Var Calculus -- Volumes of Revolution

consider the curves y = 6 x = 0 and y = x²+2

Revolve the bound area around the y-axis and find the volume of the product solid.

Here's what I did.


r = x = (y - 2)^1/2

V = pi * INTEGRAL from y = 2 -> y = 6 of r² dy = y - 2 dy = y²/2 - 2y from 6 - > 2 = 8pi


All of my calculations are correct, but I am not sure if i selected the radius of the solid correctly. This is why i chose to do this problem -- I need to get better at figuring out what the radius of the solid will be.

Did I choose it correctly? If not, what is correct, and is there any good way to check yourself to see if you did indeed chose your radius correctly?

 

[JeSuisConf]

r = x = (y - 2)^1/2

V = pi * INTEGRAL from y = 2 -> y = 6 of r² dy = y - 2 dy = y²/2 - 2y from 6 - > 2 = 8pi

I think you're on the right track but I am confused by what you are doing here. The way you typed the equation is unclear.

 

[HallsofIvy]

consider the curves y = 6 x = 0 and y = x²+2

Your lack of punctuation confused me for a moment! I was trying to figure out what "y= 6x= 0" could mean! Of course, you mean y= 6, x= 0, and $y= x^2+ 2$.

Revolve the bound area around the y-axis and find the volume of the product solid.

Here's what I did.


r = x = (y - 2)^1/2

Okay, at each z, the x- coordinate on the boundary is $x= (y- 2)^{1/2}$ and that is the radius of the disk that revolving around the y-axis produces. The area of such a disk is $A= \pi r^2= \pi (y- 2)$.

V = pi * INTEGRAL from y = 2 -> y = 6 of r² dy = y - 2 dy = y²/2 - 2y from 6 - > 2 = 8pi

You need to learn to write mathematics more carefully- don't use "= " between things that are not equal: in particular, "$\pi\int_2^6 y- 2 dy$" is NOT equal to "$y- 2 dy$"!

What you mean is that the volume is given by
[tex]\pi \int_2^6 y- 2 dy= \pi(\frac{1}{2}y^2- 2y)right|_2^6= \pi\left[\left(\frac{1}{2}(36)- 2(6)\right)- \left(\frac{1}{2}(4)- 2(2)\right)\right]= 8\pi[/itex]
so, yes, your final answer is correct.

All of my calculations are correct, but I am not sure if i selected the radius of the solid correctly. This is why i chose to do this problem -- I need to get better at figuring out what the radius of the solid will be.

Since you are rotating around the y- axis, the radius (for the disk method) is perpendicular to the y-axis and so parallel to the x-axis. Since one end of a radius is at the y-axis, the other end is on the graph and the length of the radius is the x-value of that point: from $y= x^2+ 2$, $y- 2= x^2$ and $x= \pm (y- 2)^{1/2}$. The graph of $y= x^2+ 2$ is symmetric around the y-axis so it does not matter whether you use $(y- 2)^{1/2}$ or $-(y- 2)^{1/2}$ as r (and, specifically, you will be squaring r so it is really $x^2= y- 2$ you need.

Did I choose it correctly? If not, what is correct, and is there any good way to check yourself to see if you did indeed chose your radius correctly?

Yes, what you did was correct. As for "checking", just think logically and do a lot of these to gain confidence in yourself.

 

[anonymity]

Your lack of punctuation confused me for a moment! I was trying to figure out what "y= 6x= 0" could mean! Of course, you mean y= 6, x= 0, and $y= x^2+ 2$.


Okay, at each z, the x- coordinate on the boundary is $x= (y- 2)^{1/2}$ and that is the radius of the disk that revolving around the y-axis produces. The area of such a disk is $A= \pi r^2= \pi (y- 2)$.


You need to learn to write mathematics more carefully- don't use "= " between things that are not equal: in particular, "$\pi\int_2^6 y- 2 dy$" is NOT equal to "$y- 2 dy$"!

What you mean is that the volume is given by
[tex]\pi \int_2^6 y- 2 dy= \pi(\frac{1}{2}y^2- 2y)right|_2^6= \pi\left[\left(\frac{1}{2}(36)- 2(6)\right)- \left(\frac{1}{2}(4)- 2(2)\right)\right]= 8\pi[/itex]
so, yes, your final answer is correct.



Since you are rotating around the y- axis, the radius (for the disk method) is perpendicular to the y-axis and so parallel to the x-axis. Since one end of a radius is at the y-axis, the other end is on the graph and the length of the radius is the x-value of that point: from $y= x^2+ 2$, $y- 2= x^2$ and $x= \pm (y- 2)^{1/2}$. The graph of $y= x^2+ 2$ is symmetric around the y-axis so it does not matter whether you use $(y- 2)^{1/2}$ or $-(y- 2)^{1/2}$ as r (and, specifically, you will be squaring r so it is really $x^2= y- 2$ you need.


Yes, what you did was correct. As for "checking", just think logically and do a lot of these to gain confidence in yourself.

Thanks for your answer; and sorry for my sporadic and informal approach. I tried to do it correctly with latex but I just couldn't get it to work right.