Suppose we have the following boundaries:
$x^2+y^2 \le a^2$
$x \ge 0$
$y \ge 0$
$0 \le z \le \sqrt{x^2+y^2}$
I need to calculate the volume of the figure which is bounded by these conditions. So I know that the most obvious way would be to use a triple integral with the following boundaries:
$$ \int_{0}^{a}\int_{0}^{\sqrt{a^2-x^2}}\int_{0}^{\sqrt{x^2+y^2}}{dzdydx}$$
However, after solving the first integral I get $$\int_{0}^{a}\int_{0}^{\sqrt{a^2-x^2}}{\sqrt{x^2+y^2}dydx}$$
but the next integral seems kinda hard to solve which makes me wonder if the volume should be calculated by another method (for example a substitution).
Can anyone help? Thanks!
It can be done that way, but it's hard. It's better to do it in cylindrical coordinates, in which case your integral becomes$$\int_0^{\pi/2}\int_0^a\int_0^rr\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{\pi a^3}6.$$