Calculating the volume of a cone using $\int_D f = \int_K \big[\int_{h(x)}^{g(x)} \ f(x,y) \ dy \big] \ dx$ : $x\in \mathbb{R^n}, y \in \mathbb{R}$

Question

There are various ways to calculate the volume of a cone. The one that I am trying is by evaluating the following integral $$\text{V} = \int_{x^2+y^2 \le a^2} \int_{0}^{\frac{h}{a} \sqrt{a^2-x^2-y^2}} \ 1 \ dz \ dK = \frac23 \pi a^2 h$$ which is twice as the expected one! Where do I do wrong?

PS I got $z$ by using some trigonometric; as $\dfrac{z^2}{h^2} = \dfrac{a^2-x^2-y^2}{a^2}.$ I guess that here $z$ is not like a 'normal' function of $(x,y)$ different from types like infinite paraboloid but even if so I don't know how it is related.

Answer 1

Because $\dfrac{z^2}{h^2} = \dfrac{a^2-x^2-y^2}{a^2}$ is wrong. The correct one is $\dfrac{z}{h} = \dfrac{a-\sqrt{x^2-y^2}}{a}$ which gives the correct answer $1/3 \pi a^2 h$.

Answer 2

That region over which you are integrating is no cone at all. It's the top half of an ellipsoid.

You will get the value that you're after computing the integral$$\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int_{\frac ha\sqrt{x^2+y^2}}^h1\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$Here, my cone is$$\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,x^2+y^2\leqslant a^2\wedge\frac{h^2}{a^2}(x^2+y^2)\leqslant z^2\leqslant h^2\wedge0\leqslant z\right\}.$$

Answer 3

Hint: Your dealing with the equation $\dfrac{z^2}{h^2} = \dfrac{a^2-x^2-y^2}{a^2}$ is wrong, try with $$\left(1-\dfrac{z}{h}\right)^2 = \dfrac{x^2+y^2}{a^2}.$$

Answer 4

We need 3 constants $ (a,h,z_1):$

Equation of right circular cone with $z_1$ apex shift

$$\frac{r}{z-z_1}=\frac{a}{h}$$ or squaring,

$$ r^2=x^2+y^2= (z-z_1)^2(a/h)^2.$$