# Calc 2 Homework re: volumes

Discussion in 'Homework Help' started by HCN1996, Sep 8, 2016.

1. ### HCN1996 Thread Starter New Member

Nov 29, 2015
26
0
where am I going wrong with this problem? Im almost positive im wrong but im not sure where I messed up. If someone could help tutor me for this one Id greatly appreciate it.
edit: solved

Last edited: Sep 8, 2016
2. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
You need to describe your shape better. You don't label your axes, so we have no idea which is x and which is y. Plus, this is a volume, which means there needs to be a z-axis as well. Without that information we have to guess or reverse engineer work that you, yourself, are almost positive is wrong in order to figure it out. The best guess I can make is that it is a hemisphere, but if that were the case you would (should) know the final answer to compare against yours since the volume of a sphere is very well known (and easy to look up).

Looking at your final result, you have (R²-R) which is definitely wrong since you can't subtract a linear distance from an area.

HCN1996 likes this.
3. ### HCN1996 Thread Starter New Member

Nov 29, 2015
26
0
yes, my professor only wrote the graph that I have on my paper on the board and told us to solve. It was very confusing because I am just learning the material but now I get what he was trying to say. The answer is 4piR^3/3 (as you can see I messed up on line after V =)....I agree that the information should have been more clear. Its times like these when I wonder why my professor is so against using a textbook. I will certainly be picking one up tomorrow.

4. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
4piR^3/3 is the volume of a complete sphere, which is very inconsistent with the diagram which looks like it might be for a hemisphere.

But your set up is for a complete sphere.

Your problem is that you did the following

$
\sqrt{R^2 \; - \; x^2} \; = \; R \; - \; x
$

You then made another mistake when you tried to evaluate the limits.

You need to be more careful and pay close attention to the details.

HCN1996 likes this.
5. ### HCN1996 Thread Starter New Member

Nov 29, 2015
26
0
thanks for the help, Im swamped in classes this semester so its been a long day. Im having trouble following what you are saying because of my inexperience but I think I fixed the issue. The mistakes I made were silly and I realized it after taking a little break. This is my final result.

File size:
198.3 KB
Views:
7
6. ### MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hi,

I was wondering how the radical magically disappeared

7. ### MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hello again,

It dawned on me that i dont think you had shown a good logical progression from equation to equation as to why you used some equations in your first attempt. Usually there is a logical explanation between lines of text that shows how we get to that equation or expression.
For example, we dont get from sqrt(r^2-x^2) to just r^2-x^2 without some explanation of why or how this is possible.
So if you had to show your reasoning then you would have quickly spotted the problem because there would be no logical reason for some transitions between expressions.

For example:
"In a circuit we have R=10 ohms and V=10 volts."
"The current is therefore 20 amps."

See how the second line does not follow from the first in any reasonable way?
Now lets write out what we thought was happening:
"In a circuit we have R=10 ohms and V=10 volts."
"Using Ohms Law, I=V/R=10/10=1 amps"
"The current is therefore 20 amps."
"But wait that cant be right because 10/10=1."

So now we know what we did wrong and can correct it.
If we left out the statement and the thought about Ohms Law, we might have not realized that 20 amps was incorrect. That can cause a loss of a lot of points on a test.

In geometric problems we usually have to show the physical significance of a transition from one expression to the next. For example if we calculate 1/4 the area of a circle doing that one pie slice, then we have to show that the other three are the same or at least mention it so that we have a good reason for summing four such slices to get the complete area of the circle.

8. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
@HCN1996: I noticed that you removed your original image. That is discouraged because it makes it difficult, if not impossible, for readers of the thread to follow the discussion. Remember, this is not just about helping you and only you -- in fact the only reason that most responders take the time to do so is because they know they are potentially helping lots of people for years to come because of the archival nature of a forum such as this. It is extremely beneficial to students to see the mistakes that others have made and how they were found and fixed.

@MrAl: I agree with you that textual descriptions of steps being taken are very helpful and should be included -- if for no other reason than that it falls under the heading of the proper care and feeding of homework graders.

In the case of his sqrt() problem what his intended step was

$
{$$\sqrt{R^2 \; - \; x^2}$$}^2 \; = \; R^2 \; - \; x^2
$

which really doesn't warrant any explanation. He just messed up the math on something that he thought was simple and obvious (and thus not warranting an explanation).

What would have gone a long way toward making things clear would have been a well-labeled diagram that matched the problem (especially since the actual problem statement wasn't provided) and a sketch or at least textual description of the differential volume being integrated.

9. ### MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hi again,

Yes, but like you said he removed his drawings and you happened to have commented on only the second diagram while i had commented on the first diagram he removed that went from:
sqrt(r^2-x^2)^2

to:
(r-x)^2

and i just noticed that the sqrt sign had disappeared, even though there is more wrong with it. For one, we cant loose the square root sign and still keep the square

Here is a copy of his original diagram, enhance to help readability and a smaller file size for faster downloading and less archive space, which should help other readers as well:

File size:
93.4 KB
Views:
4
10. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
I was referring to that step in the original diagram -- I remember what he had in the original diagram because I commented on that exact error in Post #4.

And he didn't go from

sqrt(r^2-x^2)^2 to sqrt(r^2-x^2)

he went from

sqrt(r^2-x^2)^2 to sqrt(r-x)

because he blew the math (as described in Post #4), not because he was trying to apply some sophisticated concept that needed to be documented. What would his documentation have been? And should he have provided similar documentation if he had gotten the math correct?

11. ### MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hello again,

Well, that is "if" there was a reasonable explanation. The first problem was the missing square root sign, and the second problem was the missing squares.
"If" there was a reasonable explanation then it should have been shown.
If it is just an error, that's a little different yes, but as i read it there always seemed to be something missing, and i think it started with the 2d diagram which doesnt explain how we ended up with a 3d formula

12. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Huh? The first problem had the square root sign:

The only difference between the two lines is what is in the red boxes, in which he mistakenly screwed the math pooch by claiming that

$
\sqrt{R^2 \; - \; x^2} \; = \; R \; - \; x
$

13. ### MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hi,

That's not the expression, and i see this as two problems not just one.
The expression is (sqrt(R^2-x^2))^2

1. The square root sign disappears even though the outside square remains.
2. The individual squares disappear

The sqrt sign can disappear only if we apply the outside square, but then the inside squares must remain:
sqrt(R^2-x^2)^2 => R^2-x^2

and not as in the original paper:
sqrt(R^2-x^2)^2 => (R-x)^2

14. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
#2 happened because, in his mind, #1 cancelled them out.

He made a single mistake -- and a very common one. He thought that the square root operator "distributes" across the expression under the radical.

He falsely generalized the relation

$
sqrt{a^2} \; = \; a
$

to

$
sqrt{a^2 \; + \; b^2 \; + \; c^2} \; = \; a \; + \; b \; + \; c
$

This is the same basic mistake that people make when they go

$
{$$a \; + \; b \; + \; c$$}^2 \; = \; a^2 \; + \; b^2 \; + \; c^2
$

I see it all the time, in both directions.