How to Find the Volume Between a Hemisphere and Cone | Triple Integral Calculus 3 Tutorial

[@LucenProject]

19:44 I could be wrong, but I don't think you HAVE to approximate the cap since you have both formulas for the sphere and the cone. Once you find that circle of intersection, you can determine its radius, which will inform its distance from the center of the sphere and its volume.

[@HighPeakEducation]

I agree. To get the full volume, you would HAVE TO approximate the cap. I think we are saying the same thing.

[@LucenProject]

​@@HighPeakEducation I'm saying we do Not have to approximate. The formula for the volume of a spherical sector is V(sector)=(2/3)*pi*h*r^2 where h is the height of the spherical cap. Subtraction of a cone won't produce the answer, but subtraction of the spherical sector will. V(sphere)=4/3)*pi*r^3 so V(hemisphere)=(2/3)*pi*r^3. We just need to find h. We can do V(solid)=V(hemisphere)-V(sector) V(solid)=(2/3)*pi*r^3 - (2/3)*pi*h*r^2 V(solid)=(2/3)*pi*(r^2)*(r-h) Since h is the height of the spherical cap it is also the radius of the sphere minus the distance between the center of the circle of intersection and the center of the sphere. We can get the circle of intersection by substituting the z^2 value of the spherical equation with the z^2 value of the cone equation. Cone equation: z=4*SQRT((x^2)+(y^2)) square both sides z^2=16*((x^2)+(y^2)) Sphere equation: (x^2)+(y^2)+(z^2)=36 substitute in 16*((x^2)+(y^2)) for z^2 (x^2)+(y^2)+16*((x^2)+(y^2))=36 group like terms 17*((x^2)+(y^2))=36 divide both sides by 17 (x^2)+(y^2)=36/17 We have the equation for the circle of intersection. A circle with the center on the z-axis and a radius^2 of 36/17. We'll call this value a^2. Now we need to find the height, h, of the cap/spherical section that sits on this circle. With the radius of this circle of intersection and the radius of the sphere as the hypotenuse, we can you use Pythagorean formula to find the last leg of the triangle they create. The height, h, will be the difference between the radius of the sphere and the remaining side of this triangle, so we'll call label the side of the triangle as (r-h). (a^2)+((r-h)^2)=r^2 subtract a^2 from both sides (r-h)^2=(r^2)-(a^2) substitute the values for r^2 and a^2 (r-h)^2=(36)-(36/17) get a common denominator (r-h)^2=(612/17)-(36/17) subtract (r-h)^2=576/17 take the SQRT of both sides r-h=24/SQRT(17) We could go on to find h specifically, but the formula for the volume of our solid from subtracting a cone from a hemisphere actually has an (r-h) term that we can substitute this in for. V(solid)=(2/3)*pi*(r^2)*(r-h) substitute the values for r^2 and (r-h) V(solid)=(2/3)*pi*(36)*(24/SQRT(17)) V(solid)=(2)*pi*(12)*(24/SQRT(17)) V(solid)=(24)*pi*(24/SQRT(17)) V(solid)=(576*pi)/SQRT(17) For a calculus course, this isn't what is wanted, but I don't think we can say it would require approximation where the other does not as the methods result in the same value.

[@HighPeakEducation]

@@LucenProject Thank you for the thorough explanation! I was not aware how to do this geometrically. Also, the solid area in this case is the hemisphere minus the snowcone, which is the cone with the spherical cap. Appreciate you adding value to this video!

[@LucenProject]

@@HighPeakEducation Glad to add. Not too long ago I was calculating the total area of intersecting circles with different radiuses if and needed the to calculate the area of the caps so I came across the spherical versions too. I find multivariable calculus intimidating, but it's nice to see there is a way to do the same there! And the tutor format is nice too! Someone else asking some of the questions I have! 😁

[@andrews476]

I respect the work, I honestly don’t know how people get a kick from doing this 😭

[@HighPeakEducation]

Thank you for watching.