Question

Calc 2 Question relating to discs and washers

Answer 1

how would i set this equation up?

Answer 2

@satellite73 @Abhisar

Answer 3

@Michele_Laino

Answer 4

from what i understand

Answer 5

@ganeshie8

Answer 6

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

Answer 7

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

Answer 8

So the square are of different sizes?

Answer 9

yes

Answer 10

wait. maybe.

Answer 11

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

Answer 12

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

Answer 13

tbh, beats me

Answer 14

i think this is more abstract

Answer 15

@hartnn @ParthKohli

Answer 16

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

Answer 17

What kind of solid is this? Clearly this solid is not radially symmetric or else you will have a contradiction.

Answer 18

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 \(\theta\).

Answer 19

\[0\leq k \leq 5r\]

Answer 20

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.

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

try \(z=2r\cosθ\)

\[4 \ \int\limits_{\theta = 0}^{\pi/2} \ \int\limits_{r = 0}^{5R} \ \int\limits_{z=0}^{2r \cos \theta} r \ dz \ dr\ d\theta\]

Answer 25

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

Answer 26

\[\frac{1000R^3}{3}\]

Answer 27

This is a hard question.

Answer 28

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

Answer 29

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.

Answer 30

\[8 \int\limits_{0}^{\pi/2} \ \int\limits_{0}^{5R} \ \sqrt{25R^2 - r^2 \sin^2 \theta}\ r\ dr\ d\theta\]

\[ = 750R^3 \left(\frac{3 \pi - 4}{6} \right)\]