How to solve this particular task about centroids












0












$begingroup$


enter image description here



Can anyone help me with 29.I know how to find centroids when one function is given but in this one I don't think that knowing only the function of circle or rectangle will help us.I found that there is something called additive centroids but I don't know how that one works here.Can anyone fully explain this task and additive centroids please?






calculus integration centroid







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Centroids are averages. Get the centroid for each piece and then take the average of their coordinates. By symmetry this amounts to taking the average of the $y$ values from each individual centroid.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:52












  • $begingroup$
    Ohh really...I did not know that, but where does it come from?I think there should be a proof for that
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 9:54










  • $begingroup$
    Look into the derivation of the centroid. It extends from a weighted average of discrete points in space to an integral.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:55










  • $begingroup$
    Honestly,I could not understand your point...
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10










  • $begingroup$
    I just want a solid proof that average can be found when we talk about unit shapes like the one given above
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10
















0












$begingroup$


enter image description here



Can anyone help me with 29.I know how to find centroids when one function is given but in this one I don't think that knowing only the function of circle or rectangle will help us.I found that there is something called additive centroids but I don't know how that one works here.Can anyone fully explain this task and additive centroids please?






calculus integration centroid







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Centroids are averages. Get the centroid for each piece and then take the average of their coordinates. By symmetry this amounts to taking the average of the $y$ values from each individual centroid.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:52












  • $begingroup$
    Ohh really...I did not know that, but where does it come from?I think there should be a proof for that
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 9:54










  • $begingroup$
    Look into the derivation of the centroid. It extends from a weighted average of discrete points in space to an integral.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:55










  • $begingroup$
    Honestly,I could not understand your point...
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10










  • $begingroup$
    I just want a solid proof that average can be found when we talk about unit shapes like the one given above
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10














0












0








0





$begingroup$


enter image description here



Can anyone help me with 29.I know how to find centroids when one function is given but in this one I don't think that knowing only the function of circle or rectangle will help us.I found that there is something called additive centroids but I don't know how that one works here.Can anyone fully explain this task and additive centroids please?






calculus integration centroid







share|cite|improve this question









$endgroup$




enter image description here



Can anyone help me with 29.I know how to find centroids when one function is given but in this one I don't think that knowing only the function of circle or rectangle will help us.I found that there is something called additive centroids but I don't know how that one works here.Can anyone fully explain this task and additive centroids please?






calculus integration centroid




calculus integration centroid






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 9:44









Arif RustamovArif Rustamov

387




387








  • 2




    $begingroup$
    Centroids are averages. Get the centroid for each piece and then take the average of their coordinates. By symmetry this amounts to taking the average of the $y$ values from each individual centroid.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:52












  • $begingroup$
    Ohh really...I did not know that, but where does it come from?I think there should be a proof for that
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 9:54










  • $begingroup$
    Look into the derivation of the centroid. It extends from a weighted average of discrete points in space to an integral.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:55










  • $begingroup$
    Honestly,I could not understand your point...
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10










  • $begingroup$
    I just want a solid proof that average can be found when we talk about unit shapes like the one given above
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10














  • 2




    $begingroup$
    Centroids are averages. Get the centroid for each piece and then take the average of their coordinates. By symmetry this amounts to taking the average of the $y$ values from each individual centroid.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:52












  • $begingroup$
    Ohh really...I did not know that, but where does it come from?I think there should be a proof for that
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 9:54










  • $begingroup$
    Look into the derivation of the centroid. It extends from a weighted average of discrete points in space to an integral.
    $endgroup$
    – John Douma
    Dec 24 '18 at 9:55










  • $begingroup$
    Honestly,I could not understand your point...
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10










  • $begingroup$
    I just want a solid proof that average can be found when we talk about unit shapes like the one given above
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:10








2




2




$begingroup$
Centroids are averages. Get the centroid for each piece and then take the average of their coordinates. By symmetry this amounts to taking the average of the $y$ values from each individual centroid.
$endgroup$
– John Douma
Dec 24 '18 at 9:52






$begingroup$
Centroids are averages. Get the centroid for each piece and then take the average of their coordinates. By symmetry this amounts to taking the average of the $y$ values from each individual centroid.
$endgroup$
– John Douma
Dec 24 '18 at 9:52














$begingroup$
Ohh really...I did not know that, but where does it come from?I think there should be a proof for that
$endgroup$
– Arif Rustamov
Dec 24 '18 at 9:54




$begingroup$
Ohh really...I did not know that, but where does it come from?I think there should be a proof for that
$endgroup$
– Arif Rustamov
Dec 24 '18 at 9:54












$begingroup$
Look into the derivation of the centroid. It extends from a weighted average of discrete points in space to an integral.
$endgroup$
– John Douma
Dec 24 '18 at 9:55




$begingroup$
Look into the derivation of the centroid. It extends from a weighted average of discrete points in space to an integral.
$endgroup$
– John Douma
Dec 24 '18 at 9:55












$begingroup$
Honestly,I could not understand your point...
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:10




$begingroup$
Honestly,I could not understand your point...
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:10












$begingroup$
I just want a solid proof that average can be found when we talk about unit shapes like the one given above
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:10




$begingroup$
I just want a solid proof that average can be found when we talk about unit shapes like the one given above
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:10










1 Answer
1






active

oldest

votes


















2












$begingroup$

As you can probably guess (by symmetry or intuition), the centroid of the rectangle is at $C_1 =(0,h/2)$. Now the centroid of semicircle is actually $frac{4r}{3pi}$ above the centre of the circle which in our case makes it



$$C_2 = (0,h+frac{4a}{3pi})$$



To calculate the centroid of the entire figure you simply use weighted averages



$$C = frac{C_1cdot 2ah + C_2cdotpi r^2}{2ah+pi r^2}$$



You can take it from here, I believe.



Reference - http://datagenetics.com/blog/january52017/index.html






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:38










  • $begingroup$
    As you see I am reading this lesson from the book and there was nothing about this term
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:39










  • $begingroup$
    A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Also could you please accept the answer then?
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Got it finally.Huge thanks
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:50












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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

As you can probably guess (by symmetry or intuition), the centroid of the rectangle is at $C_1 =(0,h/2)$. Now the centroid of semicircle is actually $frac{4r}{3pi}$ above the centre of the circle which in our case makes it



$$C_2 = (0,h+frac{4a}{3pi})$$



To calculate the centroid of the entire figure you simply use weighted averages



$$C = frac{C_1cdot 2ah + C_2cdotpi r^2}{2ah+pi r^2}$$



You can take it from here, I believe.



Reference - http://datagenetics.com/blog/january52017/index.html






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:38










  • $begingroup$
    As you see I am reading this lesson from the book and there was nothing about this term
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:39










  • $begingroup$
    A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Also could you please accept the answer then?
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Got it finally.Huge thanks
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:50
















2












$begingroup$

As you can probably guess (by symmetry or intuition), the centroid of the rectangle is at $C_1 =(0,h/2)$. Now the centroid of semicircle is actually $frac{4r}{3pi}$ above the centre of the circle which in our case makes it



$$C_2 = (0,h+frac{4a}{3pi})$$



To calculate the centroid of the entire figure you simply use weighted averages



$$C = frac{C_1cdot 2ah + C_2cdotpi r^2}{2ah+pi r^2}$$



You can take it from here, I believe.



Reference - http://datagenetics.com/blog/january52017/index.html






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:38










  • $begingroup$
    As you see I am reading this lesson from the book and there was nothing about this term
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:39










  • $begingroup$
    A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Also could you please accept the answer then?
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Got it finally.Huge thanks
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:50














2












2








2





$begingroup$

As you can probably guess (by symmetry or intuition), the centroid of the rectangle is at $C_1 =(0,h/2)$. Now the centroid of semicircle is actually $frac{4r}{3pi}$ above the centre of the circle which in our case makes it



$$C_2 = (0,h+frac{4a}{3pi})$$



To calculate the centroid of the entire figure you simply use weighted averages



$$C = frac{C_1cdot 2ah + C_2cdotpi r^2}{2ah+pi r^2}$$



You can take it from here, I believe.



Reference - http://datagenetics.com/blog/january52017/index.html






share|cite|improve this answer









$endgroup$



As you can probably guess (by symmetry or intuition), the centroid of the rectangle is at $C_1 =(0,h/2)$. Now the centroid of semicircle is actually $frac{4r}{3pi}$ above the centre of the circle which in our case makes it



$$C_2 = (0,h+frac{4a}{3pi})$$



To calculate the centroid of the entire figure you simply use weighted averages



$$C = frac{C_1cdot 2ah + C_2cdotpi r^2}{2ah+pi r^2}$$



You can take it from here, I believe.



Reference - http://datagenetics.com/blog/january52017/index.html







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 10:27









Sauhard SharmaSauhard Sharma

953318




953318












  • $begingroup$
    Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:38










  • $begingroup$
    As you see I am reading this lesson from the book and there was nothing about this term
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:39










  • $begingroup$
    A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Also could you please accept the answer then?
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Got it finally.Huge thanks
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:50


















  • $begingroup$
    Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:38










  • $begingroup$
    As you see I am reading this lesson from the book and there was nothing about this term
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:39










  • $begingroup$
    A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Also could you please accept the answer then?
    $endgroup$
    – Sauhard Sharma
    Dec 24 '18 at 10:45










  • $begingroup$
    Got it finally.Huge thanks
    $endgroup$
    – Arif Rustamov
    Dec 24 '18 at 10:50
















$begingroup$
Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:38




$begingroup$
Huge thanks for great answer.One more thing,it is the first time that I encounter term weighted averages, What does it mean in general?Can that formula be applied to all kind of united shapes?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:38












$begingroup$
As you see I am reading this lesson from the book and there was nothing about this term
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:39




$begingroup$
As you see I am reading this lesson from the book and there was nothing about this term
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:39












$begingroup$
A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
$endgroup$
– Sauhard Sharma
Dec 24 '18 at 10:45




$begingroup$
A normal or non-weighted average is where you sum everything up and divide by the total number. However, in a weighted average, every quantity has a different weight and each value is multiplied by its weight and then divided by the sum of weights. Reference - math.tutorvista.com/statistics/weighted-average.html
$endgroup$
– Sauhard Sharma
Dec 24 '18 at 10:45












$begingroup$
Also could you please accept the answer then?
$endgroup$
– Sauhard Sharma
Dec 24 '18 at 10:45




$begingroup$
Also could you please accept the answer then?
$endgroup$
– Sauhard Sharma
Dec 24 '18 at 10:45












$begingroup$
Got it finally.Huge thanks
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:50




$begingroup$
Got it finally.Huge thanks
$endgroup$
– Arif Rustamov
Dec 24 '18 at 10:50


















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