Volume of water: A conical glass is tilted against a wall. It is filled to the brim with water. Calculate the volume of water.

[mario99]

I never checked details of the OP. There's a typo here:

The 255 should be 225:

[imath]E:35x^2-30x-225+36y^2=0[/imath]​

[imath]y_{high}=\frac{1}{6}\sqrt{225+30x-35x^2}[/imath]​

The endpoints of this ellipse are indeed 3 and [imath]-\frac{15}{7}[/imath]. So it's just a typo on that one line, not an error that propagates. In fact, you can see the 225 in the work below that.

Thank you a lot Dr.Peterson.

[imath]\displaystyle \int_{-15/7}^{3}\int_{-(1/6)\sqrt{225+30x-35x^2}}^{(1/6)\sqrt{225+30x-35x^2}}\int_{(3-x)/2}^{9-3\sqrt{x^2+y^2}} \approx 51.2061[/imath]

I spent two days trying to figure out where was my mistake while the OP has given us a wrong equation. If the ellipse equation on xy-plane was not given, is there a way to find it out?

 

[Dr.Peterson]

Thank you a lot Dr.Peterson.

[imath]\displaystyle \int_{-15/7}^{3}\int_{-(1/6)\sqrt{225+30x-35x^2}}^{(1/6)\sqrt{225+30x-35x^2}}\int_{(3-x)/2}^{9-3\sqrt{x^2+y^2}} \approx 51.2061[/imath]

I spent two days trying to figure out where was my mistake while the OP has given us a wrong equation. If the ellipse equation on xy-plane was not given, is there a way to find it out?

It wasn't given; he worked it out! (And I checked that work in order to correct the typo.)

The equation of the cone is [imath]C:x^2+y^2=\left(\frac{9-z}{3}\right)^2[/imath]
Isolating z gives: [imath]z_{high}=9-3\sqrt{x^2+y^2}[/imath]

The equation of the inclined plane is [imath]V:2z+x=3[/imath]
Isolating z gives: [imath]z_{low}=\frac{3-x}{2}[/imath]

The intersection of the cone and the plane is an ellipse.
It's projection on the XY-plane is also an ellipse with equation [imath]E:35x^2-30x-[225]+36y^2=0[/imath]
Isolating y gives: [imath]y_{high}=\frac{1}{6}\sqrt{[225]+30x-35x^2}[/imath]

Plug the expression for z from the plane into the equation for the cone (that is, set the two z's equal), and you get an equation in x and y, which is as shown (corrected).

 

[Janosik]

OMG!!! I am so sorry!!!
It is indeed a typo
And as Dr.Peterson said: I used the correct value in my work...
@Dr.Peterson : thank you so much for pointing out that typo.

[imath]\displaystyle \int_{-15/7}^{3}\int_{-(1/6)\sqrt{225+30x-35x^2}}^{(1/6)\sqrt{225+30x-35x^2}}\int_{(3-x)/2}^{9-3\sqrt{x^2+y^2}} \approx 51.2061[/imath]

Can you tell me wich solver you use?

 

[mario99]

OMG!!! I am so sorry!!!
It is indeed a typo
And as Dr.Peterson said: I used the correct value in my work...
@Dr.Peterson : thank you so much for pointing out that typo.


Can you tell me wich solver you use?

Wolfram Alpha

I tried to solve the integral in Cylindrical coordinate, but I get incorrect result. I think that the Jacobian of the ellipse is not r. I looked around and I found this: when the ellipse is [imath]\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/imath], the Jacobian is [imath]abr[/imath]. But when the ellipse is shifted like our ellipse, I don't know how to find the Jacobian. Since it is not convenient to solve this integral by hand, it is better just to solve it by CAS, or other approximation methods like Simpson's Rule.

I have solved tons of cone volume problems, but this is the first time to see a problem like this. Thanks to the OP who proved that there are still a lot of things to learn even when you think that you are a professional.