anonymous
  • anonymous
Integrate f(x,y,z)=z over the domain 0<=z<=x^2+y^2<=9
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Domain is \[0 \le z \le x^2 + y^2 \le 9 \]
anonymous
  • anonymous
I am also convert everything to cylindrical coordinates before I integrate. I am unsure as to what the limits of integration should be. I am thinking that z goes from 0 to r^2 and r goes form 0 to 3
Kainui
  • Kainui
I suggest drawing out the region of integration, that'll help a lot I think in making the other limits of integration perfectly clear.

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Kainui
  • Kainui
Another helpful tip, cut up that weird thing of limits of \(0 \le z \le x^2+y^2 \le 9\) into separate pieces: \[0 \le z\]\[z \le x^2+y^2\]\[x^2+y^2 \le 9\]
anonymous
  • anonymous
Is the region a paraboloid from z=0 to z=9?
anonymous
  • anonymous
The region is indeed a paraboloid. Suppose we set \(y=0\), then \(z=x^2\) (a parabola in the \(x\)-\(z\) plane); suppose \(x=0\), then \(z=y^2\) (the same but in the \(y\)-\(z\) plane). For any value of \(0\le z\le9\), you get the equation of a circle with radius \(\sqrt z\). See the sketch (not drawn to scale):|dw:1449273351913:dw|
anonymous
  • anonymous
Here's a far better plot rendered in Mathematica
1 Attachment
anonymous
  • anonymous
So the region of integration \(D\) is given by the set of all points \((x,y,z)\) that fill up this paraboloid cup. Converting to cylindrical coordinates, you have \[\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=z\end{cases}\]which gives \[|J|=\begin{vmatrix}\dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial \theta}&\dfrac{\partial x}{\partial z}\\[1ex]\dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial \theta}&\dfrac{\partial y}{\partial z}\\[1ex]\dfrac{\partial z}{\partial r}&\dfrac{\partial z}{\partial \theta}&\dfrac{\partial z}{\partial z}\end{vmatrix}=\begin{vmatrix}\cos\theta&-r\sin\theta&0\\[1ex]\sin\theta&r\cos\theta&0\\[1ex]0&0&1\end{vmatrix}=r\]so that \[\iiint_Df(x,y,z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\iiint_DF(r,\theta,z)r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z\]In Cartesian coordinates, the limits of both \(x\) and \(y\) are determined by the radius of the circle with radius \(\sqrt z\) (green in the plot), which at most means that \(-3\le x\le3\) and \(-3\le y\le 3\). The space between the paraboloid and the plane \(z=9\) immediately give you the limits for \(z\), ie. \(x^2+y^2\le z\le 9\). The change to cylindrical coordinates gives \(x^2+y^2=r^2\), so the the limits of \(z\) are easy to find: \(r^2\le z\le9\). For any circular cross-section, \(\theta\) will have to vary over \(0\le\theta\le2\pi\). Lastly, \(r\) would need to cover every point on any ray from the origin to the paraboloid, which means that at least \(r=0\) and at most \(r=3\). So, \[\int_{-3}^3\int_{-3}^3\int_{x^2+y^2}^9z\,\mathrm{d}z\,\mathrm{d}x\,\mathrm{d}y=\int_0^{2\pi}\int_0^3\int_{r^2}^9zr\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta\]

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