Datasets:
NationalTutoringObservatory
/
MathEd-PII

Dataset card Data Studio Files
xet
 Community
1
role
stringclasses
2 values
content
stringlengths
0
2.1k
session_id
int64
10
21.7k
sequence_id
int64
0
2.38k
annotations
listlengths
0
8
volunteer
Yes!
17,825
45
[]
volunteer
Very good.
17,825
46
[]
volunteer
We can use either x or y to integrate here. We just have to choose.
17,825
47
[]
volunteer
The limits for the x must run from -r to r
17,825
48
[]
volunteer
Think of the circular base as centered at the origin.
17,825
49
[]
student
wait but how does that relate to (2y)^2 being the area??
17,825
50
[]
volunteer
That is precisely the integral you set up.
17,825
51
[]
volunteer
(2y)^2 = 4y^2 = 4(r^2-x^2)
17,825
52
[]
volunteer
Then you sum up each thin slice of an area over the limits.
17,825
53
[]
volunteer
I'm not sure I'm addressing your concern.
17,825
54
[]
student
but why is it 2y and not 2r?
17,825
55
[]
volunteer
r is a constant. Like 4, 16, etc.
17,825
56
[]
volunteer
It does not vary.
17,825
57
[]
student
why couldnt it just be y?
17,825
58
[]
volunteer
It is the radius of our given circle
17,825
59
[]
student
why can't the entire height just be y then
17,825
60
[]
student
if it vaires
17,825
61
[]
volunteer
If you imagine the view from the top and think about the center slice...
17,825
62
[]
volunteer
It is exactly 2r by 2r
17,825
63
[]
student
yea
17,825
64
[]
volunteer
But other slices are shorter
17,825
65
[]
volunteer
They can be represented by 2y by 2y.
17,825
66
[]
volunteer
That is 4y^2
17,825
67
[]
student
so are we saying r=y?
17,825
68
[]
student
and just not using r because it's a constant
17,825
69
[]
volunteer
Let's back up. I feel like I'm not answering your questions.
17,825
70
[]
student
ok
17,825
71
[]
volunteer
Think of the circle.
17,825
72
[]
volunteer
It's the at bottom of our structure.
17,825
73
[]
volunteer
It is sitting with a center at the origin and has an equation x^2 + y^2 = r^2
17,825
74
[]
student
ok
17,825
75
[]
volunteer
Now if we construct a very thin square over the origin, it would be 2r by 2r. Do you see that?
17,825
76
[]
student
yes
17,825
77
[]
volunteer
Now. This the leap of faith.
17,825
78
[]
volunteer
Each of the squares have sides that get smaller as we move away from the center.
17,825
79
[]
volunteer
Their sides are like 2r, but smaller.
17,825
80
[]
student
ok
17,825
81
[]
volunteer
The relationship is expressed as 2y by 2y.
17,825
82
[]
volunteer
That is the area of a "representative" square.
17,825
83
[]
volunteer
It's not just y^2
17,825
84
[]
student
so y is just a smaller rep version of r
17,825
85
[]
volunteer
When compared to the very center slice.
17,825
86
[]
volunteer
Yes.
17,825
87
[]
student
ok
17,825
88
[]
student
could it also be (2x)^2 if i wanted to take the integral in terms of x?
17,825
89
[]
volunteer
Yes. Then the integral would be using the y variable and dy
17,825
90
[]
volunteer
It is the same.
17,825
91
[]
student
ok
17,825
92
[]
volunteer
Now. If you buy all that..
17,825
93
[]
volunteer
We must set the limits of integration.
17,825
94
[]
student
ok i get it
17,825
95
[]
volunteer
The lower limit for x is -r and the upper limit for x is +r.
17,825
96
[]
volunteer
So. If you buy that, change a and b to -r and r
17,825
97
[]
volunteer
The reason for this is those are the smallest and larget values of x.
17,825
98
[]
volunteer
largest
17,825
99
[]
volunteer
Questions?
17,825
100
[]
volunteer
It may take a while to "sink in."
17,825
101
[]
student
no i dont that makes sense
17,825
102
[]
volunteer
Ok. Let's finish the easy part and inegrate.
17,825
103
[]
student
ok
17,825
104
[]
volunteer
integrate.
17,825
105
[]
volunteer
Let's change a and b to -r and r
17,825
106
[]
student
ok
17,825
107
[]
volunteer
Good. Want to take a shot at integrating?
17,825
108
[]
student
not really the 2 variables is throwing me off
17,825
109
[]
volunteer
Remember r is a constant.
17,825
110
[]
volunteer
There is only one variable, x
17,825
111
[]
volunteer
So the integration would be as follows:
17,825
112
[]
volunteer
4[(r^2)x - (1/3)(x^3)]
17,825
113
[]
student
isnt it 4/3
17,825
114
[]
student
not 1/3?
17,825
115
[]
volunteer
or (4r^2)(x) - (4/3)(x^3)
17,825
116
[]
student
ok
17,825
117
[]
volunteer
There are two terms to integrate.
17,825
118
[]
volunteer
The first is simply a constant.
17,825
119
[]
volunteer
So, it becomes 4r^2(x)
17,825
120
[]
volunteer
The second is as you said.
17,825
121
[]
volunteer
Ok. Now we plug in the upper limit of r to give 4r^3 - (4/3)r^3
17,825
122
[]
volunteer
From it we will subtract the lower limit evaluated at -r
17,825
123
[]
student
ok
17,825
124
[]
volunteer
Give me a minute to calculate
17,825
125
[]
student
i got this
17,825
126
[]
volunteer
Should give a final answer of (16/3)r^3
17,825
127
[]
student
oh what did I do wrong ://
17,825
128
[]
volunteer
The last term should be negative
17,825
129
[]
volunteer
8r^3 - (4/3)r^3 - (4/3)r^3
17,825
130
[]
volunteer
8r^3 - (8/3)r^3
17,825
131
[]
volunteer
(24/3)r^3 - (8/3)r^3 = (16/3)r^3
17,825
132
[]
volunteer
Just simplify what you have and you will get the same answer
17,825
133
[]
student
ok got it
17,825
134
[]
volunteer
That was a lot. You hung in there!
17,825
135
[]
volunteer
I think it will be easier to visualize as you get more experience.
17,825
136
[]
volunteer
I like that you questioned things!
17,825
137
[]
student
thanks :))
17,825
138
[]
student
it made sense thanks for explaining
17,825
139
[]
student
I do have 2 more like this though :(((
17,825
140
[]
volunteer
These are not easy!
17,825
141
[]
volunteer
Let's see another.
17,825
142
[]
student
ok :))
17,825
143
[]
volunteer
Let me think for a few minutes.
17,825
144
[]