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anonymous

  • one year ago

Identify the definite integral that computes the volume of the solid generated by revolving the region bounded by the graph of y = x^3 and the line y = x, between x = 0 and x = 1, about the y axis. Please show steps. Thanks

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  1. Michele_Laino
    • one year ago
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    here is the drawing of your question? |dw:1438789614976:dw|

  2. Michele_Laino
    • one year ago
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    so your integral is given by the subsequent expression: \[\Large \int_0^1 {dy} \pi {\left( {\sqrt[3]{y} - y} \right)^2}\]

  3. Michele_Laino
    • one year ago
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    so we have the subsequent steps: \[\Large \begin{gathered} \int_0^1 {dy} \pi {\left( {\sqrt[3]{y} - y} \right)^2} = \pi \int_0^1 {dy} \left( {{y^{2/3}} + {y^2} - {y^{4/3}}} \right) = \hfill \\ \hfill \\ = \pi \left. {\left( {\frac{3}{5}{y^{5/3}} + \frac{{{y^3}}}{3} - \frac{3}{7}{y^{7/3}}} \right)} \right|_0^1 = \hfill \\ \hfill \\ = \pi \left( {\frac{3}{5} + \frac{1}{3} - \frac{3}{7}} \right) = \frac{{8\pi }}{{105}} \hfill \\ \end{gathered} \]

  4. Astrophysics
    • one year ago
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    |dw:1438789876309:dw| \[V = \int\limits_{a}^{b} A(x) dx\] Where the Area is \[A = \pi r^2 \implies A= \pi(\sqrt[3]{y}-y)^2\]\[\int\limits_{0}^{1} \pi (\sqrt[3]{y}-y)^2 dy\] this is called the disk method :-)

  5. Michele_Laino
    • one year ago
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    oops.. I have made a typo: here are the right steps: \[\Large \begin{gathered} \int_0^1 {dy} \pi {\left( {\sqrt[3]{y} - y} \right)^2} = \pi \int_0^1 {dy} \left( {{y^{2/3}} + {y^2} - 2{y^{4/3}}} \right) = \hfill \\ \hfill \\ = \pi \left. {\left( {\frac{3}{5}{y^{5/3}} + \frac{{{y^3}}}{3} - \frac{6}{7}{y^{7/3}}} \right)} \right|_0^1 = \hfill \\ \hfill \\ = \pi \left( {\frac{3}{5} + \frac{1}{3} - \frac{6}{7}} \right) = \frac{{8\pi }}{{105}} \hfill \\ \end{gathered} \]

  6. IrishBoy123
    • one year ago
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    |dw:1438792815058:dw| \(dV \approx \pi [ \ (x+dx)^2 - x^2] [ \ x - x^3]\) \( = \pi [ \ x^2+2xdx +dx^2 - x^2] [ \ x - x^3]\) \( \approx \pi \ 2x dx (x - x^3) = 2 \pi (x^2 - x^4) \ dx\) [\(dx^2 \rightarrow 0\)] \(\therefore \large V = 2 \pi \ \int_{0}^{1} x^2 - x^4 \ dx = 2 \pi (\frac{1}{3} - \frac{1}{5}) = \frac{4 \pi}{15}\)

  7. IrishBoy123
    • one year ago
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    the other way |dw:1438793806147:dw| \(dV \approx [ \ \pi (\sqrt[3]{y})^2 - \pi y^2 \ ] dy\) \(\large V = \pi \int_{0}^{1} \ \ y^{2/3} - y^2 dy \\ = \pi [ \frac{3}{5} y^{5/3} - \frac{1}{3}y^3]_{0}^{1} \\ = \pi [ \frac{3}{5} - \frac{1}{3} ]= \frac{4 \pi}{15}\)

  8. Astrophysics
    • one year ago
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    Ahhh Michele and I made the same mistake haha, thanks @IrishBoy123

  9. Michele_Laino
    • one year ago
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    I'm very sorry @Yaros. Thanks @IrishBoy123 for your reply

  10. Astrophysics
    • one year ago
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    I sort of forgot that \[A = \pi ((\text{outer~r}^2-\text{inner r}^2))\] around those lines

  11. Michele_Laino
    • one year ago
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    yes! that's right! @Astrophysics

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