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

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1. anonymous

2. anonymous

how would i set this equation up?

3. anonymous

@satellite73 @Abhisar

4. anonymous

@Michele_Laino

5. anonymous

|dw:1441942141903:dw| from what i understand

6. anonymous

@ganeshie8

7. thomas5267

What exactly is "parallel cross section perpendicular to the base"?

8. anonymous

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

So the square are of different sizes?

10. anonymous

yes

11. anonymous

wait. maybe.

12. anonymous

no i think that's a yes. because they're changing to fit the shape of the circular disk

13. thomas5267

What is the shape of the solid? I am really confused.

14. anonymous

tbh, beats me

15. anonymous

i think this is more abstract

16. anonymous

@hartnn @ParthKohli

17. thomas5267

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

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

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

$0\leq k \leq 5r$

21. thomas5267

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

3D plot of the solid. Weird Solid 2.png looks weird because it is a top down view of 3D plot.

23. thomas5267

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

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

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

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

27. IrishBoy123

$\frac{1000R^3}{3}$

28. thomas5267

This is a hard question.

29. thomas5267

\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

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

$8 \int\limits_{0}^{\pi/2} \ \int\limits_{0}^{5R} \ \sqrt{25R^2 - r^2 \sin^2 \theta}\ r\ dr\ d\theta$ $= 750R^3 \left(\frac{3 \pi - 4}{6} \right)$