# Triple integral in Cartesian Coordinates

Let $$R$$ be the region in the first octant bounded by a surface $$F(x,y,z) = 0$$ and the coordinate planes. The projection of $$R$$

1. on the $$xy$$-plane is bounded by the coordinates axes and the curve $$y = 9 - x^2$$ ,
2. on the $$xz$$-plane is bounded by the coordinates axes and the curve $$x =\sqrt{9-z}$$.
3. on the $$yz$$-plane is bounded by the coordinates axes and the curve $$z=9-y$$.

Now, the volume of the region $$R$$ is asked. When I sketch the region, I get $$1/8$$ of a sphere cutted by the plane $$z=9-y$$. Then, I write a triple integral as follows. $$\int_{0}^{9}\int_{0}^{9-z}\int_{0}^{\sqrt{9-z}}dx \, dy \, dz$$

However, the solution says the inner integral has bounds $$\int_{0}^{\sqrt{9-y-z}} dx .$$ What is the part that I am missing?

• If I understand the region correctly, neither integrals seem correct. You will have to split the region into two sub-regions at $y = z$. Apr 23, 2022 at 15:54
• I agree with MathLover. This is a challenging question because we're not given actual equations of surfaces that determine the region. The upper bound on $x$ should be the smaller of $\sqrt{9-z}$ and $\sqrt{9-y}$. Where they got $\sqrt{9-y-z}$ is ... a mystery. Apr 24, 2022 at 0:32

Note that $$y=9-x^2$$, $$z=9-x^2$$ and $$y+z=9$$. So, the region $$R$$ is symmetric with respect to the plane $$z=y$$, which allows the volume integration to be set up as follows $$V=2\int_0^{\frac92}dz \int_z^{9-z}dy \int_0^{\sqrt{9-y}}dx$$