Find the centroid bounded by $x+y = 2, y=x^2, y=0$.
$begingroup$
I found the intersection point $(-2,4)$ and $(1,1)$
I decided to make everything in terms of $y$. Thus, $y=x^2$, $y = 2-x$.
So I will use the integral bound of -2 to 1.
$2-x$ is the higher curve, $x^2$ is the lower curve clearly from graphing
The formula for the $x$ coordinate is $frac{int_{-2}^1((2-x)-x^2)x ,dx}{int_{-2}^1((2-x)-x^2),dx}$.
We can also then proceed to find the $y$ coordinate. However my calculations are not yielding the same $x$ value as the solutions which is$(frac{52}{45},frac{20}{63})$
integration centroid
edited Jul 9 '15 at 1:13
Linus S.
1,818518
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
$endgroup$
add a comment |
$begingroup$
I found the intersection point $(-2,4)$ and $(1,1)$
I decided to make everything in terms of $y$. Thus, $y=x^2$, $y = 2-x$.
So I will use the integral bound of -2 to 1.
$2-x$ is the higher curve, $x^2$ is the lower curve clearly from graphing
The formula for the $x$ coordinate is $frac{int_{-2}^1((2-x)-x^2)x ,dx}{int_{-2}^1((2-x)-x^2),dx}$.
We can also then proceed to find the $y$ coordinate. However my calculations are not yielding the same $x$ value as the solutions which is$(frac{52}{45},frac{20}{63})$
integration centroid
edited Jul 9 '15 at 1:13
Linus S.
1,818518
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
$endgroup$
add a comment |
$begingroup$
I found the intersection point $(-2,4)$ and $(1,1)$
I decided to make everything in terms of $y$. Thus, $y=x^2$, $y = 2-x$.
So I will use the integral bound of -2 to 1.
$2-x$ is the higher curve, $x^2$ is the lower curve clearly from graphing
The formula for the $x$ coordinate is $frac{int_{-2}^1((2-x)-x^2)x ,dx}{int_{-2}^1((2-x)-x^2),dx}$.
We can also then proceed to find the $y$ coordinate. However my calculations are not yielding the same $x$ value as the solutions which is$(frac{52}{45},frac{20}{63})$
integration centroid
edited Jul 9 '15 at 1:13
Linus S.
1,818518
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
$endgroup$
I found the intersection point $(-2,4)$ and $(1,1)$
I decided to make everything in terms of $y$. Thus, $y=x^2$, $y = 2-x$.
So I will use the integral bound of -2 to 1.
$2-x$ is the higher curve, $x^2$ is the lower curve clearly from graphing
The formula for the $x$ coordinate is $frac{int_{-2}^1((2-x)-x^2)x ,dx}{int_{-2}^1((2-x)-x^2),dx}$.
We can also then proceed to find the $y$ coordinate. However my calculations are not yielding the same $x$ value as the solutions which is$(frac{52}{45},frac{20}{63})$
integration centroid
integration centroid
edited Jul 9 '15 at 1:13
Linus S.
1,818518
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
edited Jul 9 '15 at 1:13
Linus S.
1,818518
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
edited Jul 9 '15 at 1:13
Linus S.
1,818518
edited Jul 9 '15 at 1:13
Linus S.
1,818518
edited Jul 9 '15 at 1:13
Linus S.
1,818518
1,818518
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
asked Jul 9 '15 at 1:06
Jason GenovaJason Genova
104
104
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let me redo this... the problem statement is a bit confusing, so I'll re-interpret it. If you were given the problem from a book, please transcribe the problem verbatim so we can best determine what is sought.
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
$endgroup$
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
Let me redo this... the problem statement is a bit confusing, so I'll re-interpret it. If you were given the problem from a book, please transcribe the problem verbatim so we can best determine what is sought.
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
$endgroup$
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
add a comment |
$begingroup$
Let me redo this... the problem statement is a bit confusing, so I'll re-interpret it. If you were given the problem from a book, please transcribe the problem verbatim so we can best determine what is sought.
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
$endgroup$
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
add a comment |
$begingroup$
Let me redo this... the problem statement is a bit confusing, so I'll re-interpret it. If you were given the problem from a book, please transcribe the problem verbatim so we can best determine what is sought.
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
$endgroup$
Let me redo this... the problem statement is a bit confusing, so I'll re-interpret it. If you were given the problem from a book, please transcribe the problem verbatim so we can best determine what is sought.
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
edited Jul 9 '15 at 1:58
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
answered Jul 9 '15 at 1:37
David G. StorkDavid G. Stork
11k31432
11k31432
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
add a comment |
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Note the y = 0, I think you are finding the centroid of the part you didnt shade in. The small part on the positive x axis, from 0 to 2.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:41
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
Are you sure? I think your diagram is wrong it should be shaded in the region of the x axis, y = x^2 and x+y=2, from x=0 to 2. How is that area you shaded bounded by y=0.
$endgroup$
– Jason Genova
Jul 9 '15 at 1:43
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
$begingroup$
The equation $y=0$ is a horizontal line coincident at all points with the $x$ axis. You seem to be confusing the $x$ and $y$ axes with regard to that constraint.
$endgroup$
– David G. Stork
Jul 9 '15 at 1:56
add a comment |
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