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anonymous
 one year ago
Calc 2 Question relating to discs and washers
anonymous
 one year ago
Calc 2 Question relating to discs and washers

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would i set this equation up?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 @Abhisar

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441942141903:dw from what i understand

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1What exactly is "parallel cross section perpendicular to the base"?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there isn't a picture, so i'm guessing it's that the squares are parallel to each other, but perpendicular to the base.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1So the square are of different sizes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i think that's a yes. because they're changing to fit the shape of the circular disk

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1What is the shape of the solid? I am really confused.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think this is more abstract

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1So the height of the solid is given by \(f(\theta)=2\times 5r\sin(\theta)=10r\sin(\theta),\,0\leq\theta\leq\ \frac{\pi}{2}\)?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1What kind of solid is this? Clearly this solid is not radially symmetric or else you will have a contradiction. dw:1441948258709:dw

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Let \((k,\theta)\) 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 \(\theta\).

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[0\leq k \leq 5r\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Correction: The function is \(f(x)=10\sqrt{25x^2}\) if the solid is described in Cartesian coordinates and the cross section is perpendicular to x axis.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.13D plot of the solid. Weird Solid 2.png looks weird because it is a top down view of 3D plot.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Actually \(f(x)=2\sqrt{25x^2}\). So: \[ x=r \cos(\theta)\\ y=r\sin(\theta)\\ z=f(x)=2\sqrt{25r^2\cos^2(\theta)}\\ \text{Volume}=\int_0^{2\pi}\int_0^{5r}2k\sqrt{25k^2\cos^2(\theta)}\,dk\,d\theta \] Not sure whether the integral is correct or not.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I am pulling my hair out! \[ f(x)=2\sqrt{25r^2x^2}\\ x=k\cos(\theta)\\ y=k\sin(\theta)\\ z=f(x)=2\sqrt{25r^2k^2\cos^2(\theta)}\\ \text{Volume}=\int_0^{2\pi}\int_0^{5r}2k\sqrt{25r^2k^2\cos(\theta)}\,dk\,d\theta \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0try \(z=2r\cosθ\) \[4 \ \int\limits_{\theta = 0}^{\pi/2} \ \int\limits_{r = 0}^{5R} \ \int\limits_{z=0}^{2r \cos \theta} r \ dz \ dr\ d\theta\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \int_0^{2\pi}\int_0^{5r}2k\sqrt{25r^2k^2\cos^2(\theta)}\,dk \, d\theta\\ u=25r^2k^2\cos^2(\theta)\\ du=2k\cos^2(\theta)\,dk\\ \int_0^{2\pi}\int_0^{5r}\frac{1}{\cos^2(\theta)}\sqrt{u}\, du\, d\theta\\ \int_0^{2\pi}\frac{2}{3\cos^2(\theta)}u^{3/2}_{k=0}^{k=5r}\,du\,d\theta\\ \int_0^{2\pi}\frac{2}{3\cos^2(\theta)}\left(25r^2k^2\cos^2(\theta)\right)^{3/2}_0^{5r}\,d\theta\\ \int_0^{2\pi}\frac{2}{3\cos^2(\theta)}\left(\left(125r^3\right)\left(1\cos^2(\theta)\right)^{3/2}125r^3\right)\,d\theta\\ \int_0^{2\pi}\frac{2}{3\cos^2(\theta)}\left(125r^3\left\sin(\theta)\right^3125r^3\right)\,d\theta\\ \int_0^{2\pi}\frac{250}{3\cos^2(\theta)}r^3\left\sin(\theta)\right^3\,d\theta+\int_0^{2\pi}\frac{250}{3\cos^2(\theta)}r^3\,d\theta\\ \] \[ \int_0^{2\pi}\frac{250}{3\cos^2(\theta)}r^3\left\sin(\theta)\right^3\,d\theta\text{ is nonconverging. Back to the drawing board.} \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1000R^3}{3}\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1This is a hard question.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \begin{align*} &\phantom{{}={}}\int_{5r}^{5r}\int_{\sqrt{25r^2x^2}}^{\sqrt{25r^2x^2}}2\sqrt{25r^2x^2}\,dy \, dx\\ &=\int_{5r}^{5r}\left[2y\sqrt{25r^2x^2}\right]_{\sqrt{25r^2x^2}}^{\sqrt{25r^2x^2}}\, dx\\ &=2\int_{5r}^{5r}\left(25r^2x^2+\left(25r^2x^2\right)\right)\, dx\\ &=4\int_{5r}^{5r}25r^2x^2\, dx\\ &=4\left[25xr^2\frac{1}{3}x^3\right]_{5r}^{5r}\\ &=4\left[125r^3\frac{125}{3}r^3\left(125r^3+\frac{125}{3}r^3\right)\right]\\ &=\frac{2000}{3}r^3 \end{align*} \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0scope 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[8 \int\limits_{0}^{\pi/2} \ \int\limits_{0}^{5R} \ \sqrt{25R^2  r^2 \sin^2 \theta}\ r\ dr\ d\theta\] \[ = 750R^3 \left(\frac{3 \pi  4}{6} \right)\]
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