Volume of Hemisphere Puzzle












-2












$begingroup$


Here we have 3 hemispheres.
The volumes of which are;
$$V_0 = y$$
$$V_1 = 15x$$
$$V_2 = 10x$$



Find $y$ and come up with a formula for finding $y$ with any values of $V_1$ and $V_2$.



enter image description here




This is my attempt at working it out but what I have at the very end looks very complicated and I wasn't expecting something like that.



$$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
$$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
$$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



Using Pythagoras' theorem we get;
$$ (2r_0)^2 = (r_1+r_2)^2 + (r_1-r_2)^2$$
$$⇒ 4(r_0)^2 = 2(r_1)^2 + 2(r_2)^2$$
$$⇒ 4left(sqrt[3]{frac{3V_0}{2pi}}right)^2 = 2left(sqrt[3]{frac{3V_1}{2pi}}right)^2 + 2left(sqrt[3]{frac{3V_2}{2pi}}right)^2$$
$$⇒ sqrt[3]{frac{3V_0}{2pi}} = left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{1}{2}}$$
$$⇒ V_0 = left({frac{2pi}{3}}right)left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{3}{2}}$$






geometry







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  • $begingroup$
    Voting to reopen! :)
    $endgroup$
    – Blue
    Dec 4 '18 at 10:20








  • 1




    $begingroup$
    I am making the edits.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:22






  • 1




    $begingroup$
    Another MathJax tip: To get parenthesis to match the height of tall things (like fractions, and your radicals with fractions inside), use left(...right). For instance, left(frac{2pi}{3}right) to get $$left(frac{2pi}{3}right)$$
    $endgroup$
    – Blue
    Dec 4 '18 at 10:27












  • $begingroup$
    think I made those edits, wish I had just use that notation to begin with.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:48






  • 1




    $begingroup$
    I see what you are saying and it is a false assumption I have made.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 11:47
















-2












$begingroup$


Here we have 3 hemispheres.
The volumes of which are;
$$V_0 = y$$
$$V_1 = 15x$$
$$V_2 = 10x$$



Find $y$ and come up with a formula for finding $y$ with any values of $V_1$ and $V_2$.



enter image description here




This is my attempt at working it out but what I have at the very end looks very complicated and I wasn't expecting something like that.



$$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
$$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
$$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



Using Pythagoras' theorem we get;
$$ (2r_0)^2 = (r_1+r_2)^2 + (r_1-r_2)^2$$
$$⇒ 4(r_0)^2 = 2(r_1)^2 + 2(r_2)^2$$
$$⇒ 4left(sqrt[3]{frac{3V_0}{2pi}}right)^2 = 2left(sqrt[3]{frac{3V_1}{2pi}}right)^2 + 2left(sqrt[3]{frac{3V_2}{2pi}}right)^2$$
$$⇒ sqrt[3]{frac{3V_0}{2pi}} = left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{1}{2}}$$
$$⇒ V_0 = left({frac{2pi}{3}}right)left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{3}{2}}$$






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Voting to reopen! :)
    $endgroup$
    – Blue
    Dec 4 '18 at 10:20








  • 1




    $begingroup$
    I am making the edits.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:22






  • 1




    $begingroup$
    Another MathJax tip: To get parenthesis to match the height of tall things (like fractions, and your radicals with fractions inside), use left(...right). For instance, left(frac{2pi}{3}right) to get $$left(frac{2pi}{3}right)$$
    $endgroup$
    – Blue
    Dec 4 '18 at 10:27












  • $begingroup$
    think I made those edits, wish I had just use that notation to begin with.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:48






  • 1




    $begingroup$
    I see what you are saying and it is a false assumption I have made.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 11:47














-2












-2








-2





$begingroup$


Here we have 3 hemispheres.
The volumes of which are;
$$V_0 = y$$
$$V_1 = 15x$$
$$V_2 = 10x$$



Find $y$ and come up with a formula for finding $y$ with any values of $V_1$ and $V_2$.



enter image description here




This is my attempt at working it out but what I have at the very end looks very complicated and I wasn't expecting something like that.



$$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
$$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
$$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



Using Pythagoras' theorem we get;
$$ (2r_0)^2 = (r_1+r_2)^2 + (r_1-r_2)^2$$
$$⇒ 4(r_0)^2 = 2(r_1)^2 + 2(r_2)^2$$
$$⇒ 4left(sqrt[3]{frac{3V_0}{2pi}}right)^2 = 2left(sqrt[3]{frac{3V_1}{2pi}}right)^2 + 2left(sqrt[3]{frac{3V_2}{2pi}}right)^2$$
$$⇒ sqrt[3]{frac{3V_0}{2pi}} = left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{1}{2}}$$
$$⇒ V_0 = left({frac{2pi}{3}}right)left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{3}{2}}$$






geometry







share|cite|improve this question











$endgroup$




Here we have 3 hemispheres.
The volumes of which are;
$$V_0 = y$$
$$V_1 = 15x$$
$$V_2 = 10x$$



Find $y$ and come up with a formula for finding $y$ with any values of $V_1$ and $V_2$.



enter image description here




This is my attempt at working it out but what I have at the very end looks very complicated and I wasn't expecting something like that.



$$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
$$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
$$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



Using Pythagoras' theorem we get;
$$ (2r_0)^2 = (r_1+r_2)^2 + (r_1-r_2)^2$$
$$⇒ 4(r_0)^2 = 2(r_1)^2 + 2(r_2)^2$$
$$⇒ 4left(sqrt[3]{frac{3V_0}{2pi}}right)^2 = 2left(sqrt[3]{frac{3V_1}{2pi}}right)^2 + 2left(sqrt[3]{frac{3V_2}{2pi}}right)^2$$
$$⇒ sqrt[3]{frac{3V_0}{2pi}} = left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{1}{2}}$$
$$⇒ V_0 = left({frac{2pi}{3}}right)left({frac{left({frac{3V_1}{2pi}}right)^{frac{2}{3}} + left({frac{3V_2}{2pi}}right)^{frac{2}{3}}}{2}}right)^{frac{3}{2}}$$






geometry




geometry






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edited Dec 4 '18 at 11:21







Ben Franks

















asked Dec 3 '18 at 9:41









Ben FranksBen Franks

261110




261110












  • $begingroup$
    Voting to reopen! :)
    $endgroup$
    – Blue
    Dec 4 '18 at 10:20








  • 1




    $begingroup$
    I am making the edits.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:22






  • 1




    $begingroup$
    Another MathJax tip: To get parenthesis to match the height of tall things (like fractions, and your radicals with fractions inside), use left(...right). For instance, left(frac{2pi}{3}right) to get $$left(frac{2pi}{3}right)$$
    $endgroup$
    – Blue
    Dec 4 '18 at 10:27












  • $begingroup$
    think I made those edits, wish I had just use that notation to begin with.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:48






  • 1




    $begingroup$
    I see what you are saying and it is a false assumption I have made.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 11:47


















  • $begingroup$
    Voting to reopen! :)
    $endgroup$
    – Blue
    Dec 4 '18 at 10:20








  • 1




    $begingroup$
    I am making the edits.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:22






  • 1




    $begingroup$
    Another MathJax tip: To get parenthesis to match the height of tall things (like fractions, and your radicals with fractions inside), use left(...right). For instance, left(frac{2pi}{3}right) to get $$left(frac{2pi}{3}right)$$
    $endgroup$
    – Blue
    Dec 4 '18 at 10:27












  • $begingroup$
    think I made those edits, wish I had just use that notation to begin with.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 10:48






  • 1




    $begingroup$
    I see what you are saying and it is a false assumption I have made.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 11:47
















$begingroup$
Voting to reopen! :)
$endgroup$
– Blue
Dec 4 '18 at 10:20






$begingroup$
Voting to reopen! :)
$endgroup$
– Blue
Dec 4 '18 at 10:20






1




1




$begingroup$
I am making the edits.
$endgroup$
– Ben Franks
Dec 4 '18 at 10:22




$begingroup$
I am making the edits.
$endgroup$
– Ben Franks
Dec 4 '18 at 10:22




1




1




$begingroup$
Another MathJax tip: To get parenthesis to match the height of tall things (like fractions, and your radicals with fractions inside), use left(...right). For instance, left(frac{2pi}{3}right) to get $$left(frac{2pi}{3}right)$$
$endgroup$
– Blue
Dec 4 '18 at 10:27






$begingroup$
Another MathJax tip: To get parenthesis to match the height of tall things (like fractions, and your radicals with fractions inside), use left(...right). For instance, left(frac{2pi}{3}right) to get $$left(frac{2pi}{3}right)$$
$endgroup$
– Blue
Dec 4 '18 at 10:27














$begingroup$
think I made those edits, wish I had just use that notation to begin with.
$endgroup$
– Ben Franks
Dec 4 '18 at 10:48




$begingroup$
think I made those edits, wish I had just use that notation to begin with.
$endgroup$
– Ben Franks
Dec 4 '18 at 10:48




1




1




$begingroup$
I see what you are saying and it is a false assumption I have made.
$endgroup$
– Ben Franks
Dec 4 '18 at 11:47




$begingroup$
I see what you are saying and it is a false assumption I have made.
$endgroup$
– Ben Franks
Dec 4 '18 at 11:47










2 Answers
2






active

oldest

votes


















1












$begingroup$

You can use the given relations on $V_1$ and $V_2$ to get their respective radii. Having found the radii in terms of $x$, you need to find the length of the direct common chord between them. Draw the radii from the the centres $C_1$ and $C_2$ to the tangent $T$, note that they are perpendicular. Let them meet the tangent at $T_1$ and $T_2$ respectively. Drop a perpendicular from $C_2$ to $C_1T_1$, let them meet at $X$. Then $C_1C_2X$ is a right angled triangle, $C_2XT_1T_2$ is a rectangle. Now use pythagorus' theorem to find the length of the tangent, which is the same as the diameter of the topmost hemisphere and equate with the area.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have given this a go and posted an answer, think I got the hang of it.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 12:08



















0












$begingroup$

$$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
$$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
$$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



Using Pythagoras' theorem we get;
$$(r_1+r_2)^2 = (2r_0)^2 + (r_1-r_2)^2$$
$$⇒(r_0)^2 = r_1r_2$$
$$⇒left({frac{3V_0}{2pi}}right)^{frac{2}{3}} = {frac{3V_1}{2pi}}^{frac{1}{3}}{frac{3V_2}{2pi}}^{frac{1}{3}}$$
$$⇒left({frac{3V_0}{2pi}}right)^2 = {frac{9V_1V_2}{4pi^2}}$$
$$⇒V_0^2 = V_1V_2$$



Now just sub in $V_1$ and $V_2$ respectively.






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    2 Answers
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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You can use the given relations on $V_1$ and $V_2$ to get their respective radii. Having found the radii in terms of $x$, you need to find the length of the direct common chord between them. Draw the radii from the the centres $C_1$ and $C_2$ to the tangent $T$, note that they are perpendicular. Let them meet the tangent at $T_1$ and $T_2$ respectively. Drop a perpendicular from $C_2$ to $C_1T_1$, let them meet at $X$. Then $C_1C_2X$ is a right angled triangle, $C_2XT_1T_2$ is a rectangle. Now use pythagorus' theorem to find the length of the tangent, which is the same as the diameter of the topmost hemisphere and equate with the area.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I have given this a go and posted an answer, think I got the hang of it.
      $endgroup$
      – Ben Franks
      Dec 4 '18 at 12:08
















    1












    $begingroup$

    You can use the given relations on $V_1$ and $V_2$ to get their respective radii. Having found the radii in terms of $x$, you need to find the length of the direct common chord between them. Draw the radii from the the centres $C_1$ and $C_2$ to the tangent $T$, note that they are perpendicular. Let them meet the tangent at $T_1$ and $T_2$ respectively. Drop a perpendicular from $C_2$ to $C_1T_1$, let them meet at $X$. Then $C_1C_2X$ is a right angled triangle, $C_2XT_1T_2$ is a rectangle. Now use pythagorus' theorem to find the length of the tangent, which is the same as the diameter of the topmost hemisphere and equate with the area.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I have given this a go and posted an answer, think I got the hang of it.
      $endgroup$
      – Ben Franks
      Dec 4 '18 at 12:08














    1












    1








    1





    $begingroup$

    You can use the given relations on $V_1$ and $V_2$ to get their respective radii. Having found the radii in terms of $x$, you need to find the length of the direct common chord between them. Draw the radii from the the centres $C_1$ and $C_2$ to the tangent $T$, note that they are perpendicular. Let them meet the tangent at $T_1$ and $T_2$ respectively. Drop a perpendicular from $C_2$ to $C_1T_1$, let them meet at $X$. Then $C_1C_2X$ is a right angled triangle, $C_2XT_1T_2$ is a rectangle. Now use pythagorus' theorem to find the length of the tangent, which is the same as the diameter of the topmost hemisphere and equate with the area.






    share|cite|improve this answer









    $endgroup$



    You can use the given relations on $V_1$ and $V_2$ to get their respective radii. Having found the radii in terms of $x$, you need to find the length of the direct common chord between them. Draw the radii from the the centres $C_1$ and $C_2$ to the tangent $T$, note that they are perpendicular. Let them meet the tangent at $T_1$ and $T_2$ respectively. Drop a perpendicular from $C_2$ to $C_1T_1$, let them meet at $X$. Then $C_1C_2X$ is a right angled triangle, $C_2XT_1T_2$ is a rectangle. Now use pythagorus' theorem to find the length of the tangent, which is the same as the diameter of the topmost hemisphere and equate with the area.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 9:46









    BoshuBoshu

    705315




    705315












    • $begingroup$
      I have given this a go and posted an answer, think I got the hang of it.
      $endgroup$
      – Ben Franks
      Dec 4 '18 at 12:08


















    • $begingroup$
      I have given this a go and posted an answer, think I got the hang of it.
      $endgroup$
      – Ben Franks
      Dec 4 '18 at 12:08
















    $begingroup$
    I have given this a go and posted an answer, think I got the hang of it.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 12:08




    $begingroup$
    I have given this a go and posted an answer, think I got the hang of it.
    $endgroup$
    – Ben Franks
    Dec 4 '18 at 12:08











    0












    $begingroup$

    $$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
    $$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
    $$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



    Using Pythagoras' theorem we get;
    $$(r_1+r_2)^2 = (2r_0)^2 + (r_1-r_2)^2$$
    $$⇒(r_0)^2 = r_1r_2$$
    $$⇒left({frac{3V_0}{2pi}}right)^{frac{2}{3}} = {frac{3V_1}{2pi}}^{frac{1}{3}}{frac{3V_2}{2pi}}^{frac{1}{3}}$$
    $$⇒left({frac{3V_0}{2pi}}right)^2 = {frac{9V_1V_2}{4pi^2}}$$
    $$⇒V_0^2 = V_1V_2$$



    Now just sub in $V_1$ and $V_2$ respectively.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
      $$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
      $$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



      Using Pythagoras' theorem we get;
      $$(r_1+r_2)^2 = (2r_0)^2 + (r_1-r_2)^2$$
      $$⇒(r_0)^2 = r_1r_2$$
      $$⇒left({frac{3V_0}{2pi}}right)^{frac{2}{3}} = {frac{3V_1}{2pi}}^{frac{1}{3}}{frac{3V_2}{2pi}}^{frac{1}{3}}$$
      $$⇒left({frac{3V_0}{2pi}}right)^2 = {frac{9V_1V_2}{4pi^2}}$$
      $$⇒V_0^2 = V_1V_2$$



      Now just sub in $V_1$ and $V_2$ respectively.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
        $$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
        $$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



        Using Pythagoras' theorem we get;
        $$(r_1+r_2)^2 = (2r_0)^2 + (r_1-r_2)^2$$
        $$⇒(r_0)^2 = r_1r_2$$
        $$⇒left({frac{3V_0}{2pi}}right)^{frac{2}{3}} = {frac{3V_1}{2pi}}^{frac{1}{3}}{frac{3V_2}{2pi}}^{frac{1}{3}}$$
        $$⇒left({frac{3V_0}{2pi}}right)^2 = {frac{9V_1V_2}{4pi^2}}$$
        $$⇒V_0^2 = V_1V_2$$



        Now just sub in $V_1$ and $V_2$ respectively.






        share|cite|improve this answer











        $endgroup$



        $$V_0 = left(frac{2π}{3}right)(r_0)^3 ∴ r_0 = sqrt[3]{frac{3V_0}{2pi}}$$
        $$V_1 = left(frac{2π}{3}right)(r_1)^3 ∴ r_1 = sqrt[3]{frac{3V_1}{2pi}}$$
        $$V_2 = left(frac{2π}{3}right)(r_2)^3 ∴ r_2 = sqrt[3]{frac{3V_2}{2pi}}$$



        Using Pythagoras' theorem we get;
        $$(r_1+r_2)^2 = (2r_0)^2 + (r_1-r_2)^2$$
        $$⇒(r_0)^2 = r_1r_2$$
        $$⇒left({frac{3V_0}{2pi}}right)^{frac{2}{3}} = {frac{3V_1}{2pi}}^{frac{1}{3}}{frac{3V_2}{2pi}}^{frac{1}{3}}$$
        $$⇒left({frac{3V_0}{2pi}}right)^2 = {frac{9V_1V_2}{4pi^2}}$$
        $$⇒V_0^2 = V_1V_2$$



        Now just sub in $V_1$ and $V_2$ respectively.







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        edited Dec 5 '18 at 0:34

























        answered Dec 4 '18 at 12:06









        Ben FranksBen Franks

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