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anonymous

  • one year ago

Calc 2 Question relating to discs and washers

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    how would i set this equation up?

  3. anonymous
    • one year ago
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    @satellite73 @Abhisar

  4. anonymous
    • one year ago
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    @Michele_Laino

  5. anonymous
    • one year ago
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    |dw:1441942141903:dw| from what i understand

  6. anonymous
    • one year ago
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    @ganeshie8

  7. thomas5267
    • one year ago
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    What exactly is "parallel cross section perpendicular to the base"?

  8. anonymous
    • one year ago
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    there isn't a picture, so i'm guessing it's that the squares are parallel to each other, but perpendicular to the base.

  9. thomas5267
    • one year ago
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    So the square are of different sizes?

  10. anonymous
    • one year ago
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    yes

  11. anonymous
    • one year ago
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    wait. maybe.

  12. anonymous
    • one year ago
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    no i think that's a yes. because they're changing to fit the shape of the circular disk

  13. thomas5267
    • one year ago
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    What is the shape of the solid? I am really confused.

  14. anonymous
    • one year ago
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    tbh, beats me

  15. anonymous
    • one year ago
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    i think this is more abstract

  16. anonymous
    • one year ago
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    @hartnn @ParthKohli

  17. thomas5267
    • one year ago
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    So the height of the solid is given by \(f(\theta)=2\times 5r\sin(\theta)=10r\sin(\theta),\,0\leq\theta\leq\ \frac{\pi}{2}\)?

  18. thomas5267
    • one year ago
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    What kind of solid is this? Clearly this solid is not radially symmetric or else you will have a contradiction. |dw:1441948258709:dw|

  19. thomas5267
    • one year ago
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    Let \((k,\theta)\) be a point on the closed disk in polar coordinate. If we choose the square cross section to be perpendicular to the x axis, then the height of the solid is independent of k and only dependent on \(\theta\).

  20. thomas5267
    • one year ago
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    \[0\leq k \leq 5r\]

  21. thomas5267
    • one year ago
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    Correction: The function is \(f(x)=10\sqrt{25-x^2}\) if the solid is described in Cartesian coordinates and the cross section is perpendicular to x axis.

  22. thomas5267
    • one year ago
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    3D plot of the solid. Weird Solid 2.png looks weird because it is a top down view of 3D plot.

  23. thomas5267
    • one year ago
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    Actually \(f(x)=2\sqrt{25-x^2}\). So: \[ x=r \cos(\theta)\\ y=r\sin(\theta)\\ z=f(x)=2\sqrt{25-r^2\cos^2(\theta)}\\ \text{Volume}=\int_0^{2\pi}\int_0^{5r}2k\sqrt{25-k^2\cos^2(\theta)}\,dk\,d\theta \] Not sure whether the integral is correct or not.

  24. thomas5267
    • one year ago
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    I am pulling my hair out! \[ f(x)=2\sqrt{25r^2-x^2}\\ x=k\cos(\theta)\\ y=k\sin(\theta)\\ z=f(x)=2\sqrt{25r^2-k^2\cos^2(\theta)}\\ \text{Volume}=\int_0^{2\pi}\int_0^{5r}2k\sqrt{25r^2-k^2\cos(\theta)}\,dk\,d\theta \]

  25. IrishBoy123
    • one year ago
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    try \(z=2r\cosθ\) \[4 \ \int\limits_{\theta = 0}^{\pi/2} \ \int\limits_{r = 0}^{5R} \ \int\limits_{z=0}^{2r \cos \theta} r \ dz \ dr\ d\theta\]

  26. thomas5267
    • one year ago
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    \[ \int_0^{2\pi}\int_0^{5r}2k\sqrt{25r^2-k^2\cos^2(\theta)}\,dk \, d\theta\\ u=25r^2-k^2\cos^2(\theta)\\ du=-2k\cos^2(\theta)\,dk\\ \int_0^{2\pi}\int_0^{5r}-\frac{1}{\cos^2(\theta)}\sqrt{u}\, du\, d\theta\\ \int_0^{2\pi}\frac{-2}{3\cos^2(\theta)}u^{3/2}|_{k=0}^{k=5r}\,du\,d\theta\\ \int_0^{2\pi}\frac{-2}{3\cos^2(\theta)}\left(25r^2-k^2\cos^2(\theta)\right)^{3/2}|_0^{5r}\,d\theta\\ \int_0^{2\pi}\frac{-2}{3\cos^2(\theta)}\left(\left(125r^3\right)\left(1-\cos^2(\theta)\right)^{3/2}-125r^3\right)\,d\theta\\ \int_0^{2\pi}\frac{-2}{3\cos^2(\theta)}\left(125r^3\left|\sin(\theta)\right|^3-125r^3\right)\,d\theta\\ \int_0^{2\pi}\frac{-250}{3\cos^2(\theta)}r^3\left|\sin(\theta)\right|^3\,d\theta+\int_0^{2\pi}\frac{250}{3\cos^2(\theta)}r^3\,d\theta\\ \] \[ \int_0^{2\pi}\frac{-250}{3\cos^2(\theta)}r^3\left|\sin(\theta)\right|^3\,d\theta\text{ is non-converging. Back to the drawing board.} \]

  27. IrishBoy123
    • one year ago
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    \[\frac{1000R^3}{3}\]

  28. thomas5267
    • one year ago
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    This is a hard question.

  29. thomas5267
    • one year ago
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    \[ \begin{align*} &\phantom{{}={}}\int_{-5r}^{5r}\int_{-\sqrt{25r^2-x^2}}^{\sqrt{25r^2-x^2}}2\sqrt{25r^2-x^2}\,dy \, dx\\ &=\int_{-5r}^{5r}\left[2y\sqrt{25r^2-x^2}\right]_{-\sqrt{25r^2-x^2}}^{\sqrt{25r^2-x^2}}\, dx\\ &=2\int_{-5r}^{5r}\left(25r^2-x^2+\left(25r^2-x^2\right)\right)\, dx\\ &=4\int_{-5r}^{5r}25r^2-x^2\, dx\\ &=4\left[25xr^2-\frac{1}{3}x^3\right]_{-5r}^{5r}\\ &=4\left[125r^3-\frac{125}{3}r^3-\left(-125r^3+\frac{125}{3}r^3\right)\right]\\ &=\frac{2000}{3}r^3 \end{align*} \]

  30. IrishBoy123
    • one year ago
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    scope it out first. we use 5R for the radius of the base because we use r in polar if the solid were a cube, it would have volume \(10R \times 10R \times 10R = 1000R^3\). that's the upper limit . if it were a cylinder, it would have volume \(\pi (5R)^2 \times 10R =\pi 250 R^3 \approx 785 R^3\) were it a cone, it would have volume \(\pi r^2 \frac{h}{3}\) , meaning \(\pi (5R)^2 . \frac{10R}{3} \approx 261 R^3\) the only way you can maintain that square cross section is by being somewhere in between.

  31. IrishBoy123
    • one year ago
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    \[8 \int\limits_{0}^{\pi/2} \ \int\limits_{0}^{5R} \ \sqrt{25R^2 - r^2 \sin^2 \theta}\ r\ dr\ d\theta\] \[ = 750R^3 \left(\frac{3 \pi - 4}{6} \right)\]

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