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anon Thu 28 May 07:00:12 2020 No.11725699 View Reply Original Report

Quoted By: >>11725760 >>11725783 >>11727151

someone please drop the answers, for some reason all of my answers seem to be incorrect and i don’t know what i’m doing wrong so if someone can just send me the answer then that would mean the world to me

Anonymous

Anonymous Thu 28 May 2020 07:38:09 No.11725760 Report

Quoted By: >>11725763 >>11725764 >>11725768 >>11727049 >>11727307

>>11725699
A sphere os described by the equation $x^2+y^2+z^2=R^2$ in $\mathbb{R} ^3$ , where $R$ is the radious. To calculate the volume we simply need to evaluate the integral:
$\iiint_S dz\,dy\,dx$ (where $S$ is the sphere). To do this it is easier to switch to spherical coordinates, so the integral becomes after calculating the Jacobian):
$\iiint_S\rho^2\sin(\phi)\,d\rho\,d\theta\,d\phi=\int_0^{\pi}\int_0^{2\pi}\int_0^R\rho^2\sin(\phi)\,d\rho\,d\theta\,d\phi=\int_0^{\pi}\int_0^{2\pi}\frac{R^3\sin(\phi)}{3}\,d\theta\,d\phi=\int_0^{\pi}\frac{2\pi R^3\sin(\phi)}{3}\,d\phi=\frac{4\pi R^3\sin(\phi)}{3}$
which you can use to get your results by pluging the radii. Also note that I didn't state it but in spherical coordinates a spher of radious R is described by the coordinates being between the limits I used in the integral.

Anonymous

Anonymous Thu 28 May 2020 07:41:41 No.11725763 Report

Quoted By:

>>11725760
last equality is wrong

Anonymous

Anonymous Thu 28 May 2020 07:41:48 No.11725764 Report

Quoted By:

>>11725760
Sorry the end result should be $\frac{4\pi R^3}{3}$ , I accidentally left the sin in by copy-pasting it.

Anonymous

Anonymous Thu 28 May 2020 07:44:32 No.11725768 Report

Quoted By: >>11726351

>>11725760
Nice. Now do it again, without spherical coordinates.

Anonymous

Anonymous Thu 28 May 2020 07:52:54 No.11725783 Report

Quoted By: >>11727246 >>11727275

>>11725699
1a) $\frac{500\pi}{3} \mathrm{cm}^3 \approx 523.6 \mathrm{cm}^3$
1b) $\frac{9\pi}{2} \mathrm{ft}^3 \approx 14.14 \mathrm{ft}^3$

The absolute state of /sci/.

Anonymous

Anonymous Thu 28 May 2020 10:45:22 No.11726138 Report

Quoted By:

4/3 • ? •r•3

This is literally it. You have to be retarded to not be able to just plug in the radius

Anonymous

Anonymous Thu 28 May 2020 12:53:19 No.11726351 Report

Quoted By: >>11727049 >>11727286

>>11725768
Ok, we can solve the equation for z:
$z=(R^2-x^2-y^2)^{\frac{1}{2}}$ or $z=-(R^2-x^2-y^2)^{\frac{1}{2}}$ so the integral:
$\iint_C(R^2-x^2-y^2)^{\frac{1}{2}}-(-(R^2-x^2-y^2)^{\frac{1}{2}})\,dy\,dx=\iint_C2(R^2-x^2-y^2)^{\frac{1}{2}}\,dy\,dx$ where C is the circle $y^2+x^2=R^2$ (we view it as a space bounded by two functions). So now you can do polar coordinates (the Jacobian is $\rho$ ) and the integral becomes:
$\iint_C2(R^2-\rho^2)^{\frac{1}{2}}\rho\,d\rho\,d\theta=2\int_0^{2\pi}\int_0^R(R^2-\rho^2)^{\frac{1}{2}}\rho\,d\rho\,d\theta=2\int_0^{2\pi}-\frac{1}{3}(R^2)^{\frac{3}{2}}\,d\theta=-2\frac{2\pi}{3}(R^3)=-\frac{4\pi R^3}{3}$
Volume is obviously positive so it's the absolute value of that... or I made a mistake with the negative somewhere.

Anonymous

Anonymous Thu 28 May 2020 18:11:42 No.11727049 Report

Quoted By:

>>11725760
>>11726351
doing gods work anon

Anonymous

Anonymous Thu 28 May 2020 18:42:55 No.11727151 Report

Quoted By: >>11727257 >>11728384

>>11725699
1a) Divide the sphere into tiny pyramids of height r from the center
V=?1/3B*h = 1/3 ? R * (R^2 d?) = 4/3 ? R^3 = 4/3 ?(125cm^3)
1b) Divide the sphere into rings of diameter D
V=???D = ??dD?D*dD = ?/2?D^2*dD = ?/6 D^3 = 27ft^3 * ?/6 = 9?/2 ft^3

Anonymous

Anonymous Thu 28 May 2020 19:16:14 No.11727246 Report

Quoted By:

>>11725783
Feeding it to him will make OP weaker, you're literally killing him anon.

Anonymous

Anonymous Thu 28 May 2020 19:20:15 No.11727257 Report

Quoted By:

>>11727151
retard in a pyramid the length of each side decreases at a linear rate as they approach the top. this is not the same, this is not two pyramids taped together, this is a sphere

Anonymous

Anonymous Thu 28 May 2020 19:25:51 No.11727275 Report

Quoted By:

>>11725783
Sig figs

Anonymous

Anonymous Thu 28 May 2020 19:27:28 No.11727286 Report

Quoted By: >>11727303

>>11726351
Why didn't you just use spherical coordinates?

Anonymous

Anonymous Thu 28 May 2020 19:31:25 No.11727303 Report

Quoted By: >>11727307

>>11727286
Are you dense anon? 11725760

Anonymous

Anonymous Thu 28 May 2020 19:32:27 No.11727307 Report

Quoted By:

>>11727303
>>11725760

Anonymous

Anonymous Fri 29 May 2020 01:01:48 No.11728384 Report

Quoted By:

>>11727151
>V=???D = ??dD?D*dD = ?/2?D^2*dD = ?/6 D^3 = 27ft^3 * ?/6 = 9?/2 ft^3

This is wrong but also correct

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