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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
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here is the drawing of your question? dw:1438789614976:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so your integral is given by the subsequent expression: \[\Large \int_0^1 {dy} \pi {\left( {\sqrt[3]{y}  y} \right)^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we have the subsequent steps: \[\Large \begin{gathered} \int_0^1 {dy} \pi {\left( {\sqrt[3]{y}  y} \right)^2} = \pi \int_0^1 {dy} \left( {{y^{2/3}} + {y^2}  {y^{4/3}}} \right) = \hfill \\ \hfill \\ = \pi \left. {\left( {\frac{3}{5}{y^{5/3}} + \frac{{{y^3}}}{3}  \frac{3}{7}{y^{7/3}}} \right)} \right_0^1 = \hfill \\ \hfill \\ = \pi \left( {\frac{3}{5} + \frac{1}{3}  \frac{3}{7}} \right) = \frac{{8\pi }}{{105}} \hfill \\ \end{gathered} \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3dw: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 :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops.. I have made a typo: here are the right steps: \[\Large \begin{gathered} \int_0^1 {dy} \pi {\left( {\sqrt[3]{y}  y} \right)^2} = \pi \int_0^1 {dy} \left( {{y^{2/3}} + {y^2}  2{y^{4/3}}} \right) = \hfill \\ \hfill \\ = \pi \left. {\left( {\frac{3}{5}{y^{5/3}} + \frac{{{y^3}}}{3}  \frac{6}{7}{y^{7/3}}} \right)} \right_0^1 = \hfill \\ \hfill \\ = \pi \left( {\frac{3}{5} + \frac{1}{3}  \frac{6}{7}} \right) = \frac{{8\pi }}{{105}} \hfill \\ \end{gathered} \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438792815058:dw \(dV \approx \pi [ \ (x+dx)^2  x^2] [ \ x  x^3]\) \( = \pi [ \ x^2+2xdx +dx^2  x^2] [ \ x  x^3]\) \( \approx \pi \ 2x dx (x  x^3) = 2 \pi (x^2  x^4) \ dx\) [\(dx^2 \rightarrow 0\)] \(\therefore \large V = 2 \pi \ \int_{0}^{1} x^2  x^4 \ dx = 2 \pi (\frac{1}{3}  \frac{1}{5}) = \frac{4 \pi}{15}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the other way dw:1438793806147:dw \(dV \approx [ \ \pi (\sqrt[3]{y})^2  \pi y^2 \ ] dy\) \(\large V = \pi \int_{0}^{1} \ \ y^{2/3}  y^2 dy \\ = \pi [ \frac{3}{5} y^{5/3}  \frac{1}{3}y^3]_{0}^{1} \\ = \pi [ \frac{3}{5}  \frac{1}{3} ]= \frac{4 \pi}{15}\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Ahhh Michele and I made the same mistake haha, thanks @IrishBoy123

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I'm very sorry @Yaros. Thanks @IrishBoy123 for your reply

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3I sort of forgot that \[A = \pi ((\text{outer~r}^2\text{inner r}^2))\] around those lines

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! that's right! @Astrophysics
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