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







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












$begingroup$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023839%2fvolume-of-hemisphere-puzzle%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 '18 at 0:34

























        answered Dec 4 '18 at 12:06









        Ben FranksBen Franks

        261110




        261110






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023839%2fvolume-of-hemisphere-puzzle%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How do I know what Microsoft account the skydrive app is syncing to?

            Image
            up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
            Read more

            Grease: Live!

            Image
            A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
            Read more

            When does type information flow backwards in C++?

            Image
            up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
            Read more