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I need to convert the following integral to spherical coordinates

$\displaystyle \int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}}\int_{0}^{x^2 + y^2} y^2 dz dy dx$

My main issue is with limits of $z$.

Using limits of $x$ and $y$, I know we need to consider the upper half of the circle $x^2 + y^2 =1$

Now, $z = x^2 + y^2$ is a paraboloid opening upwards,cut off by plane $z =1$

So, by this logic the limits for $z$ should be:

$x^2 + y^2 \le z \le 1$,

I don't get how the limits of $z$ are between $ 0$ and $x^2 + y^2$,can someone please clear this confusion for me ?

Thank you.

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  • $\begingroup$ I don't see a $z=1$ plane. I see a $z=0$ plane. $\endgroup$ – Ninad Munshi May 20 at 5:48
  • $\begingroup$ What prevents $z$ to be limited by $0$ and $x^2+y^2$? And there is no plane $z=1$ in the problem. $\endgroup$ – user May 20 at 5:48
  • $\begingroup$ it is not clear how this integral simplifies using spherical coordinates, try instead cylindrical coordinates $\endgroup$ – Masacroso May 20 at 5:48
  • $\begingroup$ @Masacroso If I correctly understood the task is to convert integral rather than to evaluate it. $\endgroup$ – user May 20 at 5:51
  • $\begingroup$ @User: I cannot understand the geometry of this integral, $z = x^2 + y^2$ is a paraboloid opening upwards ,if $ 0 < z < x^2 + y^2$, I can't understand how the limits of $x$ and $y$ calculated ? $\endgroup$ – sat091 May 20 at 5:55
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Sketch for solution: as the integral is defined you have that $$ 0\leqslant z\leqslant x^2+y^2,\quad 0\leqslant y^2\leqslant 1-x^2,\quad 0\leqslant x^2\leqslant 1\tag1 $$ The spherical coordinates are given by $$ x:=r\cos \alpha \sin \beta ,\quad y:=r \sin \alpha \sin \beta ,\quad z:=r\cos \beta \\ \text{ for }\alpha \in [0,2\pi ),\quad \beta \in [0,\pi ),\quad r\in [0,\infty )\tag2 $$ Therefore $(1)$ becomes $$ 0\leqslant r\cos \beta \leqslant r^2\sin^2 \beta ,\quad 0\leqslant r^2\sin^2 \beta \leqslant 1,\quad 0\leqslant r^2\cos^2 \alpha \sin^2 \beta \leqslant 1\tag3 $$ what simplifies to $$ 0\leqslant r\cos \beta \leqslant r^2\sin ^2\beta \leqslant 1\tag4 $$ What remains is to find the range of valid values for $r, \alpha $ and $\beta $ from $(4)$ and it defining bounds (stated on $(2)$), and rewrite the integral accordingly.

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  • $\begingroup$ But we are integrating over upper half of circle , $0 < y <\sqrt{1-x^2}$, so $\alpha$ should be $0$ to $\pi$, is this incorrect ? $\endgroup$ – sat091 May 20 at 6:34
  • $\begingroup$ It seems like that @sat091... $\endgroup$ – Anton Vrdoljak May 20 at 6:46
  • $\begingroup$ @sat091 yes, you are right... the original condition $0\le y\le\sqrt{1-x^2}$ gives $\sin\alpha \ge 0$ so $\alpha\in[0,\pi]$ $\endgroup$ – Masacroso May 20 at 9:48
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The limits of the rightmost integral are $0$ and $x^2+y^2$. Clearly $0\le x^2+y^2$ for any real $x$ and $y$. Therefore the limits are well-defined and $0\le z\le x^2+y^2$. The lateral surface of the body are the plane $y=0$ and the cylinder $x^2+y^2=1$ (for $y\ge0$).

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