#Need help pls Integral
116 messages · Page 1 of 1 (latest)
<@&286206848099549185> PLS
So we have a sphere with 4cm radius and a hole into the sphere was drilled. The hole has a radius of 1cm. All right
What formula do you know for the volume of a sphere?
So you actually have to integrate
exactly
ok
do I get your agreement sir?
no me wait
This is what I think is going on
The blue one is the ball and the red is the whole
subtracting both from each other gives us basically a new ball with radius h
ok
What I don't get is it says h² = R² - r²
and I though it would be h = R - r
adonhs
yes
and x^2=15 and r^2=1
Its what I have to do
yes
ok...
Look what I did
yeah my neck broke
I just dont know what is h+ and -h
,rotate
well let's ignore our skepticsm
it says h² = R²-r²
ok
then it follows $h = \sqrt{R^2-r^2}$
adonhs
yes
ok so we dont have the y
no so we dont have the x
yes
it just doesnt make sense to me
try something
Ok I read your French Solution so they substituted
why didnt you tell me
$\int_{y=-h}^{y=+h} \pi(R^2-y^2-r^2)dy$
what do u mean
adonhs
On remplace bien sur x² = R² - y²
are u french
where did u find the solution
$\int_{y=-\sqrt{R^2-r^2}}^{y=+\sqrt{R^2-r^2}} \pi(R^2-y^2-r^2)dy$
adonhs
I stole it from your teacher

Can you integrate this it's simple
I dont have a teacher
ok main character
ok so like I said the solution is \int _{-\sqrt{15}}^{\sqrt{15}}:\pi \left(\left(4^2-y^2\right)-1^2\right)dy
my friend its the same thing
and its equal 20\sqrt{15}\pi
$I = \int_{y=-\sqrt{15}}^{y=+\sqrt{15}} \pi(16-y^2-1)dy = \pi \int_{y=-\sqrt{15}}^{y=+\sqrt{15}}(15-y^2)dy \ \= 15\pi y - \frac{\pi}{3}y^3 \bigg |_{-\sqrt{15}}^{\sqrt{15}}$
adonhs
Is there a mistake?
its 20 But ty sir
thats y im asking
$=15\pi \sqrt{15}- \frac{\pi}{3}(\sqrt{15})^3 - 15\pi (-\sqrt{15}) + \frac{\pi}{3}(-\sqrt{15})^3$
adonhs
$=20\pi\sqrt{15}$
adonhs
I found the mistake
there was a minus instead of plus
$=20\pi\sqrt{15} \approx 243.346$
adonhs
Exactly thank you sir
,w V = 4/3pi4^3-pi1^2*8
bonjour XD
oh i meant merci