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
\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]<br /> so, yes, your final answer is correct. <br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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. </div> </div> </blockquote> 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.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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? </div> </div> </blockquote> Yes, what you did was correct. As for &quot;checking&quot;, 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
\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]<br /> so, yes, your final answer is correct. <br /> <br /> <br /> <br /> 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.<br /> <br /> <br /> Yes, what you did was correct. As for &quot;checking&quot;, just think logically and do a lot of these to gain confidence in yourself.

<br /> <br /> <br /> Thanks for your answer; and sorry for my sporadic and informal approach. I tried to do it correctly with latex but I just couldn&#039;t get it to work right.