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    I am stuck on these questions. Can someone please help me
    The region enclosed between the graphs of y=x and y=x^2 is denotated by R. Find the volume generated when R is rotated through 360° about
    a) the x axis
    b) the y axis
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    Have you tried anything?

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    Also tagging Sonechka Lemur14 to move this to the right section.
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    (Original post by _gcx)
    Have you tried anything?

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    Also tagging Sonechka Lemur14 to move this to the right section.

    the curve and the line cross at the origin but how can you go further with only one integral?
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    (Original post by tania62)
    the curve and the line cross at the origin but how can you go further with only one integral?
    Where else do they cross? (what's the other solution to x^2 = x?) A sketch might help. On this sketch label R. (if your confusion lies in having to subtract two integrals, there's no problem there)
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    (Original post by tania62)
    I am stuck on these questions. Can someone please help me
    The region enclosed between the graphs of y=x and y=x^2 is denotated by R. Find the volume generated when R is rotated through 360° about
    a) the x axis
    b) the y axis
    If you sketch the two equations y=x and y=x^2 on a single graph, you'll notice that the region enclosed by them is given by \displaystyle \int_0^1 x .dx - \int_0^1 x^2 .dx. Now if you imagine rotating these on your graph 360 degrees around the x-axis, you'd notice that the 3D region enclosed by both is given similarly where it is (rotation of x around x-axis)-(rotation of x^2 around the x-axis)

    Also note how the former is simply a cone of length \sqrt{2} and radius 1.

    Then attempt the rotation of your region around the y-axis.
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    (Original post by _gcx)
    Where else do they cross? (what's the other solution to x^2 = x?) A sketch might help. On this sketch label R. (if your confusion lies in having to subtract two integrals, there's no problem there)
    Thank youuu. I got my answer to be 2/15pi
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    (Original post by tania62)
    Thank youuu. I got my answer to be 2/15pi
    yup, have you done part b as well?
 
 
 
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