Question

Hi guys, can anyone help me with this triple integral? Many thanks:)

Answer 1

Another triple integral.  We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero.  That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$  

I gotta go so I'll stop here, sorry.

Answer 2

Converting to spherical coordinates makes the task easier:

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Under this transformation, the integrand reduces to

$x^2+xy+y^2=(1+\cos\theta\sin\theta)\rho^2\sin^2\varphi$

and the integral is

$\displaystyle\iiint_Kx^2+xy+y^2\,\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^\pi\int_0^{2\pi}\int_0^2(1+\cos\theta\sin\theta)\rho^4\sin^3\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

You can separate the variables easily in this case, so you can consider the integrals with respect to the individual variables:

$\displaystyle\int_0^\pi\sin^3\varphi\,\mathrm d\varphi=\frac43$

$\displaystyle\int_0^{2\pi}1+\cos\theta\sin\theta\,\mathrm d\theta=2\pi$

$\displaystyle\int_0^2\rho^4\,\mathrm d\rho=\frac{32}5$

Then the triple integral has a value equal to the product of these three integrals, so

$\displaystyle\iiint_Kx^2+xy+y^2\,\mathrm dx\,\mathrm dy\,\mathrm dz=\boxed{\frac{256\pi}{15}}$