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anonymous

  • 4 years ago

the region bounded by y=x2, y=1, and y-axis.find the solid obtain by rotating the region about the line y=-2

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  1. amistre64
    • 4 years ago
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    |dw:1327845541729:dw|

  2. amistre64
    • 4 years ago
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    lets rotate the pic to see it better; flip it about the y=x line

  3. amistre64
    • 4 years ago
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    |dw:1327845638231:dw|

  4. amistre64
    • 4 years ago
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    if we shell this; our radius moves from 2 to 3 but our height only takes values from 0 to 1; so we really should shift our height to match the radius values. x = sqrt(y-2) will shift the inputs appropriately

  5. amistre64
    • 4 years ago
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    so, since the height of each shell is a function of y: h(y), and our radius depend on the value of y: \[Area_{shell}=2\pi(y)(\sqrt{y-2})\] looks to be our integrand; from 2 to 3 \[2pi\int_{2}^{3}y\sqrt{y-2}dy\] but we should also see this from the "washer" method to see if it might be easier to do

  6. amistre64
    • 4 years ago
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    |dw:1327846324731:dw|

  7. amistre64
    • 4 years ago
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    Define R ro be the outside radius, and r to be the inside radius for any given section; then the area of the washer that we are concerned with is then: \[pi(R^2)-pi(r^2)=pi(R^2-r^2)\] since R is always going to be 3 and r depends on the function itself: sqrt(y-2) this becomes:\[Area_{washer}=pi(3^2-(\sqrt{y-2})^2)=pi(9-(y-2))\]

  8. amistre64
    • 4 years ago
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    r depends on y=x^2 actually: x = sqrt(y-2) x^2 = y-2 x^2+2 = y \[pi\int_{0}^{1}(3^2-(x^2+2)^2)dx\] \[pi\int_{0}^{1}(9-(x^4+4x^2+4))dx\] \[pi\int_{0}^{1}(9-x^4-4x^2-4)dx\] \[pi\int_{0}^{1}(5-x^4-4x^2)dx\] with any luck :)

  9. amistre64
    • 4 years ago
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    that looks like an easier integration, and the wolf tells me they are the same results, so lets try this washer one :)

  10. amistre64
    • 4 years ago
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    \[pi(5-\frac{1}{5}-\frac{4}{3})-pi(5(0)\cancel{-\frac{1}{5}(0)^4-\frac{4}{3}(0)^3)}^{\ 0}\] \[(5-\frac{23}{15})pi\] \[(\frac{75-23}{15})pi\] \[\frac{52}{15}pi\]

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