# Converting an Integral to Spherical Coordinates

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.

• I don't see a $z=1$ plane. I see a $z=0$ plane. – Ninad Munshi May 20 at 5:48
• What prevents $z$ to be limited by $0$ and $x^2+y^2$? And there is no plane $z=1$ in the problem. – user May 20 at 5:48
• it is not clear how this integral simplifies using spherical coordinates, try instead cylindrical coordinates – Masacroso May 20 at 5:48
• @Masacroso If I correctly understood the task is to convert integral rather than to evaluate it. – user May 20 at 5:51
• @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 ? – sat091 May 20 at 5:55

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.
• 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 ? – sat091 May 20 at 6:34
• @sat091 yes, you are right... the original condition $0\le y\le\sqrt{1-x^2}$ gives $\sin\alpha \ge 0$ so $\alpha\in[0,\pi]$ – Masacroso May 20 at 9:48
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 $$y\ge0$$).