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I know it's best to draw a picture and figure it out intuitively. But I was wondering if there was a more formal method for determining bounds of integration on triple integrals in spherical and cylindrical coordinates. For example,say I wanted to evaluate the following integral in spherical coordinates

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

How could I figure out the bounds on the triple integral in spherical coordinates WITHOUT drawing a picture?

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    $\begingroup$ Good question I've been wondering this for a long time too $\endgroup$
    – Babu
    Nov 9, 2020 at 18:50
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    $\begingroup$ Yes you have to consider the limits of integration to figure out which region you are integrating over. In this case we know that the region we are integrating over is the first octant of a unit sphere. $\endgroup$
    – Math Lover
    Nov 9, 2020 at 19:21

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Ultimately, you want to work out a system of inequalities without drawing anything. That may be difficult to accomplish even in one dimension.

Following your example, you want to work out

$$\left\{ \begin{array}{ll} 0 \leq x \leq 1\\ 0 \leq y \leq \sqrt{1-x^2}\\ 0 \leq z \leq \sqrt{1-x^2-y^2}\\ \end{array} \right. \: = \: \left\{ \begin{array}{ll} 0 \leq \rho \cos{\theta} \cos{\varphi} \leq 1\\ 0 \leq \rho \sin{\theta} \cos{\varphi} \leq \sqrt{1 -\rho^2 \cos^2{\theta} \cos^2{\phi}}\\ 0 \leq \rho \sin{\varphi} \leq \sqrt{1-\rho^2 \cos^2{\phi}}\\ \end{array} \right.$$

into

$$\left\{ \begin{array}{ll} 0 \leq \rho \leq 1\\ 0 \leq \theta \leq 90°\\ 0 \leq \varphi \leq 90°\\ \end{array} \right.$$

knowing that

$$\left\{ \begin{array}{ll} 0 \leq \rho < \infty\\ -180° \leq \theta \leq +180°\\ 0 \leq \varphi \leq 90°\\ \end{array} \right.$$

unambiguously maps

$$\left\{ \begin{array}{ll} -\infty < x < \infty\\ -\infty < y < \infty\\ -\infty < z < \infty\\ \end{array} \right. \: = \: \left\{ \begin{array}{ll} -\infty < \rho \cos{\theta} \cos{\varphi} < \infty\\ -\infty < \rho \sin{\theta} \cos{\varphi} < \infty\\ -\infty < \rho \sin{\varphi} < \infty\\ \end{array} \right.$$.

If you want to be even more rigorous, you may treat the spherical coordinate system as a mere change of variables. In that case you would have to work out the unambiguous mapping in the first place.

I don't think it's worth a try, but if you do, remember that there are more inequalities from sine and cosine definitions.

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