Calculate the volume of $G=\{(x,y,z) \in \Bbb R^3 : x^2+y^2+z^2 \leq 16 , 0 \leq z \leq 2 \}$

Question

Calculate the volume of $G=\{(x,y,z) \in \Bbb R^3 : x^2+y^2+z^2 \leq 16 , 0 \leq z \leq 2 \}$

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since they ask for volume we need $$\iiint_v1\,dV$$

in the solution in the book they used spherical coordinates I am trying something else to see if it is possible

at first I got $z=\pm \sqrt{16-x^2-y^2}$ and I also put $z=0$ and $z=2$ and got that $x^2+y^2=16$ and $x^2+y^2=12$

I assumed this is the chnage in the radius on $D$ when I calculate the double integral so $\sqrt{12} \leq r \leq 4$ and no boundaries on $\theta$ so $0 \leq \theta \leq 2 \pi$

I thought that I have everything here but the solution is not correct

$$\iint_D\left[\int_{z=-\sqrt{16-x^2-y^2}}^{{z=\sqrt{16-x^2-y^2}}}1\,dz\right]$$

$$\iint2 \sqrt{16-r^2} \cdot r\,dr\,d\theta = \frac{16}{3}\cdot 2 \pi= \frac{32 \pi}{3} \quad( \sqrt{12} \leq r \leq 4 , 0 \leq \theta \leq 2 \pi)$$

which is not correct , the correct answer is $$\frac{88 \pi}{3}$$

what am I missing? is it even possible the way I am trying?

I am aware that I could have just done spherical coordinates as they did in the book but I want another way

thanks for any help and tips

Answer 1

Basic geometry + triple integral which is easily done as double integral in polar coordinates.

You want the volume out of the upper half of the canonical sphere with radius $\;4\;$, the part of it between the "floor" (the plane $\;z=0\;$) and the "ceiling" determined by the plane $\;z=2\;$.

Volume of half the canonical sphere of radius $\;4\;=\frac12\left(\frac43\pi\cdot4^3\right)=\frac{128\pi}3\;$

Now, the plane $\;z=2\;$ cuts the half sphere forming a body for which

$$-\sqrt{12}\le x\le\sqrt{12}\;,\;\;-\sqrt{12-x^2}\le y\le\sqrt{12-x^2}\;,\;\;2\le z\le\sqrt{16-x^2-y^2}$$

and we do a triple integral for this body's volume, which becomes a sum of two easy double integrals:

$$\int_{-\sqrt{12}}^\sqrt{12}\int_{-\sqrt{x^2-12}}^{\sqrt{x^2-12}}\int_2^\sqrt{16-x^2-y^2} dz\,dy\,dx=\int_{-\sqrt{12}}^\sqrt{12}\int_{-\sqrt{x^2-12}}^{\sqrt{x^2-12}}\left(\sqrt{16-x^2-y^2}-2\right) dy\,dx\stackrel{\text{polar coord.}}=$$

$$=\int_0^\sqrt{12}\int_0^{2\pi} r\left(\sqrt{16-r^2}-2\right)d\theta=2\pi\left(-\frac12\int_0^\sqrt{12}-2r\sqrt{16-r^2}\,dr-2\pi\int_0^\sqrt{12}2r\,dr\right)=$$

$$=\left.-\pi\cdot\frac23\left(16-r^2\right)^{3/2}\right|_0^\sqrt{12}-2\pi r^2|_0^{\sqrt{12}}=-\frac{2\pi}3\left(4^{3/2}-16^{3/2}\right)-24\pi=\frac{112-72}3\pi=\frac{40\pi}3$$

And now just substract the last quantity from the first one:

$$\frac{128\pi}3-\frac{40\pi}3=\frac{88\pi}3$$

Answer 2

We denote by $E(z)$ the disc of radius $\sqrt{16-z^2}$.

The volume of G is given by :

$$\int_G dxdydz = \int_{z \in [0,2]} \left ( \int_{(x,y) \in E(z)} dxdy \right) dz $$

$$= \int_{z \in [0,2]}\int_{r \in [0,\sqrt{16-z^2}]} \int_{\theta \in [0,2\pi]}rdr d \theta dz $$

$$= 2\pi \int_{z \in [0,2]}\int_{r \in [0,\sqrt{16-z^2}]} rdrdz $$

$$= \pi \int_{z \in [0,2]} (16-z^2)dz = \frac{88}{3} \pi$$

This matches the Correct Answer.

Answer 3

As commented above:

• "What am I missing?"

  1. $r$ is not necessarily $≥\sqrt{12}.$ It goes from $0$ to $4.$
  2. $z$ is $≥0.$ It cannot reach $−\sqrt{16−r^2}.$
  3. $z$ is not only $≤\sqrt{16−r^2}.$ It only goes from $0$ to $\min(\sqrt{16−r^2},2).$

• "Is it even possible the way I am trying?"

  Yes it is! Changing nothing (not even the order of integration) in your attempt, except of course these three corrections, you get the correct answer: $$2\pi\left(\int_0^{\sqrt{12}}2rdr+\int_{\sqrt{12}}^4\sqrt{16-r^2}\,rdr\right)=2\pi\left(12+\frac83\right)=\frac{88\pi}3.$$