Calculating a Triple Integral in a Bounded Region

[mathmari]

Hey!

Let $D$ be the space $\{x,y,z)\mid z\geq 0, x^2+y^2\leq 1, x^2+y^2+z^2\leq 4\}$. I want to calculate the integral $\iiint_D x^2\,dx\,dy\,dz$. I have done the following:

We have that $x^2+y^2+z^2\leq 4\Rightarrow z^2\leq 4-x^2-y^2 \Rightarrow -\sqrt{4-x^2-y^2}\leq z\leq \sqrt{4-x^2-y^2}$. Since $z\geq 0$ we get that $0\leq z \leq \sqrt{4-x^2-y^2}$.

It holds that $4-x^2-y^2\geq 0$, i.e. $x^2+y^2\leq 4$ since $x^2+y^2\leq 1$.

We have that $x^2+y^2\leq 1 \Rightarrow y^2\leq 1-x^2 \Rightarrow -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}$.

It must hold that $1-x^2\geq 0 \Rightarrow x^2\leq 1 \Rightarrow -1\leq x\leq 1$.

Therefore $D$ can be written also in the following form: $$D=\{(x,y,z)\mid -1\leq x\leq 1, -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}, 0\leq z\leq \sqrt{4-x^2-y^2}\}$$

Therefore, we get \begin{align*}\iiint_D x^2 \,dx\,dy\,dz&=\int_{-1}^1 \left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left (\int_0^{\sqrt{4-x^2-y^2}}x^2 \ dz\right ) \ dy\right ) \ dx \\ & =\int_{-1}^1 \left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2\sqrt{4-x^2-y^2} \ dy\right ) \ dx \\ & =\int_{-1}^1 x^2\left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2} \ dy\right ) \ dx \end{align*}

To calculate the inner integral $\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2}dy$ we do the following:

We set $y=\sqrt{4-x^2}\cdot \sin t$, then $dy=\sqrt{4-x^2}\cdot \cos t \ dt$.

If $y=-\sqrt{1-x^2}$ then $t=\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )$ and if $y=\sqrt{1-x^2}$ then $t=\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )$.

Therefore we get the following:
\begin{align*}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2}dy&=\int_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )}(4-x^2)\cdot \cos^2t \ dt \\ & = (4-x^2)\cdot\int_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )} \frac{1+\cos (2t)}{2} \ dt \\ & = \frac{4-x^2}{2}\cdot\left [ t+\frac{\sin (2t)}{2} \right ]_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )} \\ & = \frac{4-x^2}{2}\cdot\left [ \arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )-\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right ) \\ +\frac{1}{2}\sin \left (2\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )\right )-\frac{1}{2}\sin \left (\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )\right )\right ]\end{align*}

Is everything correct so far? How could we continue? (Wondering)

 

[Prove It]

Hey!

Let $D$ be the space $\{x,y,z)\mid z\geq 0, x^2+y^2\leq 1, x^2+y^2+z^2\leq 4\}$. I want to calculate the integral $\iiint_D x^2\,dx\,dy\,dz$. I have done the following:

We have that $x^2+y^2+z^2\leq 4\Rightarrow z^2\leq 4-x^2-y^2 \Rightarrow -\sqrt{4-x^2-y^2}\leq z\leq \sqrt{4-x^2-y^2}$. Since $z\geq 0$ we get that $0\leq z \leq \sqrt{4-x^2-y^2}$.

It holds that $4-x^2-y^2\geq 0$, i.e. $x^2+y^2\leq 4$ since $x^2+y^2\leq 1$.

We have that $x^2+y^2\leq 1 \Rightarrow y^2\leq 1-x^2 \Rightarrow -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}$.

It must hold that $1-x^2\geq 0 \Rightarrow x^2\leq 1 \Rightarrow -1\leq x\leq 1$.

Therefore $D$ can be written also in the following form: $$D=\{(x,y,z)\mid -1\leq x\leq 1, -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}, 0\leq z\leq \sqrt{4-x^2-y^2}\}$$

Therefore, we get \begin{align*}\iiint_D x^2 \,dx\,dy\,dz&=\int_{-1}^1 \left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left (\int_0^{\sqrt{4-x^2-y^2}}x^2 \ dz\right ) \ dy\right ) \ dx \\ & =\int_{-1}^1 \left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2\sqrt{4-x^2-y^2} \ dy\right ) \ dx \\ & =\int_{-1}^1 x^2\left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2} \ dy\right ) \ dx \end{align*}

To calculate the inner integral $\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2}dy$ we do the following:

We set $y=\sqrt{4-x^2}\cdot \sin t$, then $dy=\sqrt{4-x^2}\cdot \cos t \ dt$.

If $y=-\sqrt{1-x^2}$ then $t=\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )$ and if $y=\sqrt{1-x^2}$ then $t=\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )$.

Therefore we get the following:
\begin{align*}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2}dy&=\int_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )}(4-x^2)\cdot \cos^2t \ dt \\ & = (4-x^2)\cdot\int_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )} \frac{1+\cos (2t)}{2} \ dt \\ & = \frac{4-x^2}{2}\cdot\left [ t+\frac{\sin (2t)}{2} \right ]_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )} \\ & = \frac{4-x^2}{2}\cdot\left [ \arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )-\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right ) \\ +\frac{1}{2}\sin \left (2\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )\right )-\frac{1}{2}\sin \left (\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )\right )\right ]\end{align*}

Is everything correct so far? How could we continue? (Wondering)

This would be infinity times easier to do with cylindrical polar co-ordinates, considering most of your boundary is a cylinder...

Since $\displaystyle \begin{align*} x = r\cos{ \left( \theta \right) } , \, y = r\sin{ \left( \theta \right) } \end{align*}$ and $\displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}$, you have

You have $\displaystyle \begin{align*} 0 \leq z \leq \sqrt{ 4 - \left( x^2 - y^2 \right) } \implies 0 \leq z \leq \sqrt{ 4 - r^2 } \end{align*}$ with $\displaystyle \begin{align*} 0 \leq r \leq 1 \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\,\pi \end{align*}$, giving

$\displaystyle \begin{align*} \int{\int{\int_D{x^2 \,\mathrm{d}x}\,\mathrm{d}y}\,\mathrm{d}z} &= \int_0^{2\,\pi}{\int_0^1{\int_0^{\sqrt{4-r^2}}{\left[ r\cos{ \left( \theta \right) } \right] ^2 \, r\,\mathrm{d}z}\,\mathrm{d}r}\,\mathrm{d}\theta} \\ &= \int_0^{2\,\pi}{ \int_0^1{ r^3\cos^2{ \left( \theta \right) } \, \left[ z \right] _0^{\sqrt{4 - r^2}} \,\mathrm{d}r } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \int_0^1{ r^3\,\sqrt{4 - r^2} \cos^2{ \left( \theta \right) } \,\mathrm{d}r } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2} \int_1^0{r^2\,\sqrt{4 - r^2} \left( -2\,r \right) \cos^2{ \left( \theta \right) } \,\mathrm{d}r } \,\mathrm{d}\theta } \end{align*}$

Now let $\displaystyle \begin{align*} u = 4 - r^2 \implies \mathrm{d}u = -2\,r \end{align*}$ noting that $\displaystyle \begin{align*} u \left( 1 \right) = 3 \end{align*}$ and $\displaystyle \begin{align*} u \left( 0 \right) = 4 \end{align*}$ giving

$\displaystyle \begin{align*} \int_0^{2\,pi}{\frac{1}{2} \int_1^0{ r^2\,\sqrt{4 - r^2} \left( -2\,r \right) \cos^2{ \left( \theta \right) } \,\mathrm{d}r }\,\mathrm{d}\theta} &= \int_0^{2\,\pi}{ \frac{1}{2}\cos^2{ \left( \theta \right) } \int_3^4{ \left( 4 - u \right) \,\sqrt{u} \,\mathrm{d}u } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2}\cos^2{ \left( \theta \right) } \int_3^4{ \left( 4\,u^{\frac{1}{2}} - u^{\frac{3}{2}} \right) \,\mathrm{d}u } \,\mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{ \frac{1}{2} \cos^2{ \left( \theta \right) } \left[ \frac{8\,u^{\frac{3}{2}}}{3} - \frac{2\,u^{\frac{5}{2}}}{5} \right] _3^4 \, \mathrm{d}\theta } \\ &= \int_0^{2\,\pi}{\frac{1}{2}\cos^2{ \left( \theta \right) } \left[ \left( \frac{64}{3} - \frac{64}{5} \right) - \left( 8\cdot 3^{\frac{1}{2}} - 2 \cdot 3^{ \frac{3}{2}} \right) \right] \,\mathrm{d}\theta} \\ &= \int_0^{2\,\pi}{ \frac{1}{2} \cos^2{ \left( \theta \right) } \left( \frac{320}{15} - \frac{192}{15} - \frac{30\,\sqrt{3}}{15} \right) \,\mathrm{d}\theta } \\ &= \frac{64 - 15\,\sqrt{3}}{15} \int_0^{2\,\pi}{ \cos^2{ \left( \theta \right) } \,\mathrm{d}\theta } \\ &= \frac{64 - 15\,\sqrt{3}}{30} \int_0^{2\,\pi}{ \left[ 1 + \cos{ \left( 2\,\theta \right) } \right] \,\mathrm{d}\theta } \\ &= \frac{64 - 15\,\sqrt{3}}{30} \left[ \theta + \frac{1}{2}\sin{ \left( 2\,\theta \right) } \right] _0^{2\,\pi} \\ &= \frac{64 - 15\,\sqrt{3}}{30} \left\{ \left[ 2\,\pi - \frac{1}{2} \sin{ \left( 4\,\pi \right) } \right] - \left[ 0 - \frac{1}{2} \sin{ \left( 0 \right) } \right] \right\} \\ &= \frac{ \left( 64 - 15\,\sqrt{3} \right) \pi }{15} \end{align*}$

 

[mathmari]

I see! Thank you very much! (Smile)