Find The Volume Between The Cone And The Sphere

How do I find the volume of a cone inscribed in a sphere when I'm given the sphere's radius and cone's height without using calculus?

To make it easier, disregard the third dimension first and treat it as an isosceles triangle inscribed in a circle, thus:[math]h = r + a [/math][math]a = h – r[/math]abr forms a right angled triangle so by Pythagoras[math]r^2 = a^2 + b^2[/math][math]b^2 = r^2 – a^2[/math]Substituting [math]h – r[/math] for [math]a[/math][math]b^2 = r^2 – (h – r)^2[/math][math]b^2 = r^2 – (h^2 – 2hr + r^2)[/math][math]b^2 = r^2 – h^2 + 2hr – r^2[/math][math]b^2 = 2hr – h^2[/math]Back into three dimensions. b is also the radius of the base of the cone.Its area is [math]\pi * b^2 [/math][math]\pi(2hr – h^2)[/math]The volume of the cone is one third of the base area multiplied by the height or[math]\displaystyle \frac{\pi(2h^2r – h^3)}{3}[/math]

A sphere and a cone are made such that their volumes are the same, but the total surface area of the cone is k times that of the sphere, where k is determined so that there is a unique cone satisfying this property. What's k?

Suppose the radius of the cone is [math]1[/math] and its height is [math]h[/math] and the radius of the sphere is [math]R[/math].Then [math]h = 4{R^3}[/math].Also, [math]4{R^2}k = 1 + \sqrt {{h^2} + 1} = 1 + \sqrt {16{R^6} + 1} [/math].Squaring and collecting terms,[math]2{R^4} - 2{R^2}{k^2} + k = 0[/math].But this is a quadratic in [math]R^2[/math] and the uniqueness condition forces the quadratic to have but one root. So[math]4{k^4} - 8k = 0[/math] and then [math]k=0[/math] or [math]k = \sqrt[3]{2}[/math]

The volume of a cone is equal to the volume of a sphere and the diameter of the sphere is equal to the height of the cone. What is the ratio of the radii of the cone and the sphere?

The volume of a cone is equal to the volume of a sphere and the diameter of the sphere is equal to the height of the cone. What is the ratio of the radii of the cone and the sphere?Before answering your question, I want to start off with a fascinating visual…[math]\qquad \text{Fig 1: Cone + Sphere = Cylinder}[/math] [1]In figure 1, [math]R[/math] is the radius of the sphere, and of the bases of the cone and cylinder. And [math]2R[/math] is the height of the cone and cylinder. The volume of the cone plus the volume of the sphere equals the volume of the cylinder. If you look at any horizontal cross section, you’ll see right away why this is true.[math]\qquad \text{Fig 2: Cross Sections of Cone, Sphere, and Cylinder}[/math] [2]Let [math]R[/math] be the radius of the sphere, and of the base of the cone. Consider the circles formed by taking a horizontal cross section at distance [math]h[/math] above the center of the sphere and of the cone.The radius of the first circle — the cone’s cross section — is simply [math]h,[/math] so its area is [math]\pi h^2.[/math]The radius of the second circle — the sphere’s cross section — is [math]\sqrt{R^2-h^2},[/math] so its area is [math]\pi\left(R^2-h^2\right).[/math]So the areas of these two circles add up to [math]\pi R^2.[/math]Interestingly, the volume of the cone is exactly one third the volume of the cylinder, which means the volume of the sphere is exactly two thirds the volume of the cylinder, and double the volume of the cone.Now, back to your question.In your question, the volumes of the cone and sphere are equal, which means we need to “scale up” the cone’s volume —and hence the area of its bases — by a factor of [math]2.[/math] So the cone’s radius needs to scale up by a factor of [math]\sqrt2.[/math] So the ratio of the cone’s radius to the sphere’s radius is [math]\sqrt2\text{:}1.[/math]I should point out (before anyone comments) that the following two cones have equal volume, as long as their base radius and height are the same:The only reason I didn’t use the right-hand cone shape in the “fascinating visual” portion of my answer is that the whole cross-section area thing wouldn’t have worked nearly so well.Footnotes[1] Classic - GeoGebra[2] Graphing Calculator - GeoGebra

How do I find the volume between the cone y=sqrt (x^2+z^2) and the sphere x^2+y^2+z^2=49?

So, the region of intersection looks something like the figure shown here.Now, we use the fact that solid angle subtended by this cone, in the sphere is given by [math]\Omega=2\pi(1-cos(\theta)), [/math] where [math]\theta[/math] is the half-angle of the 2D cone. That can easily be calculated from the given information - the slant height of the cone is the radius of the sphere = 7. The radius of the base of the cone is calculated by finding out the radius of the circle of intersection of the cone and sphere. That gives us the half-angle of the cone. Now we simply use the fact that the whole volume of the sphere is distributed equally in [math]4\pi [/math] steradians. So the volume enclosed by this section of the sphere must be [math]\frac{\Omega} {4\pi}\frac{4r^3}{3}[/math]