Find the volume of a region using the shell method: $x^frac23+y^frac23=4$

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A region $R$ is bounded above by the graph of $x^frac23+y^frac23=4$ and below by the x-axis. Find the volume of the region. Rotating region $R$ about the vertical line $x=8$ generates a solid of revolution $S$. I am confused with the picture below. Why is the radius considered to be $8-x$ shouldn't it be $x-8$?



shell-method




calculus integration volume




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  • For the shell drawn in the diagram above, $x$ appears to be approximately $2$. So the radius is $8-x = 8-2 = 6$. By your reasoning, the radius would be $x-8 = 2-8 = -6$.
    – Brian Tung
    Jul 24 at 17:26










  • That makes sense.
    – Jinzu
    Jul 24 at 17:28














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0
down vote

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A region $R$ is bounded above by the graph of $x^frac23+y^frac23=4$ and below by the x-axis. Find the volume of the region. Rotating region $R$ about the vertical line $x=8$ generates a solid of revolution $S$. I am confused with the picture below. Why is the radius considered to be $8-x$ shouldn't it be $x-8$?



shell-method




calculus integration volume




share|cite|improve this question



















  • For the shell drawn in the diagram above, $x$ appears to be approximately $2$. So the radius is $8-x = 8-2 = 6$. By your reasoning, the radius would be $x-8 = 2-8 = -6$.
    – Brian Tung
    Jul 24 at 17:26










  • That makes sense.
    – Jinzu
    Jul 24 at 17:28












up vote
0
down vote

favorite









up vote
0
down vote

favorite











A region $R$ is bounded above by the graph of $x^frac23+y^frac23=4$ and below by the x-axis. Find the volume of the region. Rotating region $R$ about the vertical line $x=8$ generates a solid of revolution $S$. I am confused with the picture below. Why is the radius considered to be $8-x$ shouldn't it be $x-8$?



shell-method




calculus integration volume




share|cite|improve this question











A region $R$ is bounded above by the graph of $x^frac23+y^frac23=4$ and below by the x-axis. Find the volume of the region. Rotating region $R$ about the vertical line $x=8$ generates a solid of revolution $S$. I am confused with the picture below. Why is the radius considered to be $8-x$ shouldn't it be $x-8$?



shell-method





calculus integration volume





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 24 at 17:23









Jinzu

328311




328311











  • For the shell drawn in the diagram above, $x$ appears to be approximately $2$. So the radius is $8-x = 8-2 = 6$. By your reasoning, the radius would be $x-8 = 2-8 = -6$.
    – Brian Tung
    Jul 24 at 17:26










  • That makes sense.
    – Jinzu
    Jul 24 at 17:28
















  • For the shell drawn in the diagram above, $x$ appears to be approximately $2$. So the radius is $8-x = 8-2 = 6$. By your reasoning, the radius would be $x-8 = 2-8 = -6$.
    – Brian Tung
    Jul 24 at 17:26










  • That makes sense.
    – Jinzu
    Jul 24 at 17:28















For the shell drawn in the diagram above, $x$ appears to be approximately $2$. So the radius is $8-x = 8-2 = 6$. By your reasoning, the radius would be $x-8 = 2-8 = -6$.
– Brian Tung
Jul 24 at 17:26




For the shell drawn in the diagram above, $x$ appears to be approximately $2$. So the radius is $8-x = 8-2 = 6$. By your reasoning, the radius would be $x-8 = 2-8 = -6$.
– Brian Tung
Jul 24 at 17:26












That makes sense.
– Jinzu
Jul 24 at 17:28




That makes sense.
– Jinzu
Jul 24 at 17:28










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The radius ought to be shown as $|x-8|$ or $|8-x|$ as a radius is a length quantity which cannot be negative, thus this would account for if $x<8$ or $x>8$ and give a positive value in either case.






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    1 Answer
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    1 Answer
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    active

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    active

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    up vote
    2
    down vote



    accepted










    The radius ought to be shown as $|x-8|$ or $|8-x|$ as a radius is a length quantity which cannot be negative, thus this would account for if $x<8$ or $x>8$ and give a positive value in either case.






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      up vote
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      accepted










      The radius ought to be shown as $|x-8|$ or $|8-x|$ as a radius is a length quantity which cannot be negative, thus this would account for if $x<8$ or $x>8$ and give a positive value in either case.






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        up vote
        2
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        accepted







        up vote
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        down vote



        accepted






        The radius ought to be shown as $|x-8|$ or $|8-x|$ as a radius is a length quantity which cannot be negative, thus this would account for if $x<8$ or $x>8$ and give a positive value in either case.






        share|cite|improve this answer













        The radius ought to be shown as $|x-8|$ or $|8-x|$ as a radius is a length quantity which cannot be negative, thus this would account for if $x<8$ or $x>8$ and give a positive value in either case.







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        answered Jul 24 at 18:46









        Rhys Hughes

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