What is the volume enclosed by the paraboloid $ z=x^{2}+y^{2}$ and the plane $z=10 $?

Question

I want calculate the volume enclosed by the paraboloid $ z=x^{2}+y^{2}$ and the plane $z=10,$ using double integral in cartesian coordinate system.

My approach:

Putting $ \ z=10 \ $, we get the circle $ \ x^{2}+y^{2}=10 .$ Then the volume $$V= \int_{-\sqrt {10}}^{\sqrt {10}} \int_{-\sqrt{10}}^{\sqrt{10}} (x^{2}+y^{2}-10)dxdy.$$

Is it right ? Any help is really appreciating?

Answer 1

The projection onto the $xy$ plane of your solid of interest is the disk given by $x^2+y^2 \leq 10$. As you noted the outline was $x^2+y^2=10$.

As such we let $D=\{(x,y) \in \mathbb{R}^2: x^2+y^2 \leq 10 \}$. Then we integrate the height of our solid over the region $D$ to find the volume of our solid. Because $z=10$ is the upper portion of our solid, the height is $10-(x^2+y^2)=10-x^2-y^2$.

The volume is thus,

$$V=\iint_{D} (10-x^2-y^2) dA$$

---

In rectangular coordinates first integrating over $x$ then $y$:

$$=\int_{-\sqrt{10}}^{\sqrt{10}} \int_{-\sqrt{10-y^2}}^{\sqrt{10-y^2}} (10-x^2-y^2) dx dy$$

---

In rectangular coordinates the other way:

$$=\int_{-\sqrt{10}}^{\sqrt{10}} \int_{-\sqrt{10-x^2}}^{\sqrt{10-x^2}} (10-x^2-y^2) dy dx$$

---

In polar coordinates:

$$=\int_{0}^{2\pi} \int_{0}^{\sqrt{10}} (10-r^2) r dr d\theta$$

Answer 2

For any $w\in[0,10]$, the area of the section $z=w$ is $\pi w$. By Cavalieri's principle it follows that the wanted volume is $\int_{0}^{10}\pi w\,dw = \color{red}{50\pi}$.