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

Hi guys, can anyone help me with this triple integral? Many thanks:)-example-1
User StanleyD
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2 Answers

5 votes

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
-√(4-z^2) and
+√(4-z^2)

Similarly the inner integral x goes between
\pm-√(4-y^2-z^2)

So we rewrite the integral


\displaystyle \int_(-2)^(2) \int_(-√(4-z^2))^(√(4-z^2)) \int_(-√(4-y^2-z^2))^(√(4-y^2-z^2)) (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,


\displaystyle\int_(-√(4-y^2-z^2))^(√(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=-√(4-y^2-z^2))^(z=√(4-y^2-z^2))

So the middle integral is


\displaystyle\int_(-√(4-z^2))^(√(4-z^2))2(x^2+xy+y^2)√(4-y^2-z^2) \ dy

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

User Robert Kaucher
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7.9k points
4 votes

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{(256\pi)/(15)}

User Stylesz
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8.2k points

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