Triple integral over the region between two paraboloids in cylindrical coordinates

Expert intro: This is a classic cylindrical-coordinates setup: first identify where the surfaces meet, then read the bounds off the geometry. The nice part is that the region is rotationally symmetric, so the integral simplifies a lot once $x^2+y^2$ becomes $r^2$.

Detailed walkthrough

(a) Geometry of the region

The two surfaces are

• $z=x^2+y^2$, an upward-opening paraboloid
• $z=4-x^2-y^2$, a downward-opening paraboloid

They intersect where their $z$-values are equal:

$x^2+y^2=4-x^2-y^2$

$2(x^2+y^2)=4 \quad\Rightarrow\quad x^2+y^2=2$

So the intersection curve is the circle

$r=\sqrt{2}, \qquad z=2$

because substituting $x^2+y^2=2$ into either surface gives $z=2$.

The region $R$ is the solid trapped between these two paraboloids, with bottom surface $z=r^2$ and top surface $z=4-r^2$, over the disk $0\le r\le \sqrt2$.

(b) Evaluate $\iiint_R z\,dV$

Use cylindrical coordinates:

$x=r\cos\theta,\quad y=r\sin\theta,\quad z=z,\quad dV=r\,dz\,dr\,d\theta$

The bounds are:

• $0\le \theta\le 2\pi$
• $0\le r\le \sqrt2$
• $r^2\le z\le 4-r^2$

So

= \int_0^{2\pi}\int_0^{\sqrt2}\int_{r^2}^{4-r^2} z\,r\,dz\,dr\,d\theta$$ Integrate with respect to $z$: $$\int_{r^2}^{4-r^2} z\,dz =\frac12\left[(4-r^2)^2-r^4\right]$$ Simplify: $$\frac12\bigl(16-8r^2+r^4-r^4\bigr)=8-4r^2$$ So the integral becomes $$\int_0^{2\pi}\int_0^{\sqrt2} (8-4r^2)r\,dr\,d\theta$$ $$=\int_0^{2\pi}\left[\int_0^{\sqrt2}(8r-4r^3)\,dr\right]d\theta$$ Compute the radial integral: $$\int_0^{\sqrt2}(8r-4r^3)\,dr =\left[4r^2-r^4\right]_0^{\sqrt2} =4(2)-4=4$$ Finally, $$\int_0^{2\pi}4\,d\theta=8\pi$$ Therefore, $$\boxed{\iiint_R z\,dV=8\pi}$$ The key idea is that symmetry turns the 3D problem into a clean one-variable radial integral. ### 💡 Pitfall guide A common slip is forgetting the Jacobian factor $r$ in cylindrical coordinates. Another easy mistake is reversing the top and bottom surfaces: here $z=4-r^2$ is above $z=r^2$ only for $0\le r\le\sqrt2$. ### 🔄 Real-world variant If the integrand were $1$ instead of $z$, you would be finding the volume of the same solid: $$\iiint_R 1\,dV =\int_0^{2\pi}\int_0^{\sqrt2}\int_{r^2}^{4-r^2} r\,dz\,dr\,d\theta.$$ If the integrand were $z^2$, the same bounds still apply, but the inner integral would become $\int z^2\,dz$, which gives a different radial polynomial. The geometry does not change; only the power of $z$ in the integrand does. ### 🔍 Related terms cylindrical coordinates, paraboloid, triple integral