2
$\begingroup$

I have been asked to convert the following triple integral into spherical coordinates and then evaluate.

$\int^1_0 \int^\sqrt{1-x^2}_0 \int^\sqrt{1-x^2-y^2}_\sqrt{x^2+y^2} xy\, dz\, dy\, dx$

Unless I'm reading the bounds wrong they appear to be giving a description of shape that is best integrated in cylindrical in two parts.

2 Answers 2

1

Use cylindrical coordinates, $ \int^1_0 \int^\sqrt{1-x^2}_0 \int^\sqrt{1-x^2-y^2}_\sqrt{x^2+y^2} xy\, dz\, dy\, dx = \int^{\pi/2}_0 \int^{1}_0 \int^\sqrt{1-r^2}_{r} (r\cos \theta)(r\sin \theta )\,r dz\, dr\, d\theta .$

0

First of all you need to transform x, y from Cartesian to Cylindrical coordinates: $x=r cos\theta$, $y=r cos\theta$ and $z=z$

So we get: $\int\limits_{0}^{1}\int\limits_{0}^{\pi/2}\int\limits_{r}^{\sqrt{1-r^2}}(r\cos \theta)(r\sin \theta )rdzd\theta dr=\int\limits_{0}^{1} \int \limits_{0}^{\pi/2}r^3\cos\theta \sin\theta (\sqrt{1-r^2}-r)d\theta dr=$
$\sin(\pi/2) \int \limits_{0}^{1} (r^3 \sqrt{1-r^2}-r^4)dr= \int \limits_{0}^{1}r^3 \sqrt{1-r^2}dr -\frac15$

You may need to check the limits but when you use necessary transformations you should be fine