volume between two surfaces

Question

In my textbook they solve this problem like this, Find volume between two surfaces $\quad$ $x^2+y^2-1\leq z \leq 2+\frac{2}{3}(x^2+y^2), \ \ z\geq0$

They have called the first region $D_1=x^2+y^2-1=0$, and second region $D_2=x^2+y^2=9$ and then they solve this problem with two integral $$\iint _{D_1} \left(2+\frac{2}{3}(x^2+y^2)-0\right) dxdy + \iint_{D_2} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right)dxdy$$

My question is that why they counted the first integral? and why they divided the region into two parts?

Thanks in advance

Answer 1

Please see a $2D$ sketch of the region (projection on $XZ$ or $YZ$ plane) and what each integral is finding.

What you see shaded on the left is what the first integral is finding. What you see shaded on the right is what the second integral is finding.

Since it needs to find the volume between solids above $XY$ plane ($z \geq 0$), it should have set up the first integral for $x^2 + y^2 - 1$ instead of for $(2 + \frac {2}{3}(x^2+y^2))$ to remove the volume below $XY$ plane.

So the correct integral would be -

$\displaystyle \iint _{D_1} (x^2 + y^2 - 1) \, dx \, dy + \iint_{D_2} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right) \, dx \, dy$

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (x^2 + y^2 - 1) \, dx \, dy = -\frac{\pi}{2}$

$\displaystyle \int_{-3}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right) \, dx \, dy = \frac{27 \pi}{2}$

Adding them gives you $13 \pi$ as the volume of the desired region.

EDIT:

As per the integral setup you sent, this is what the textbook is doing.

Answer 2

Projection on $Oxy$ of intersection of paraboloids $z=x^2+y^2-1$ and $z= 2+\frac{2}{3}(x^2+y^2)$ is $x^2+y^2=9$, so first integral is volume of cylinder with projection $x^2+y^2=1$ and "roof" $z= 2+\frac{2}{3}(x^2+y^2)$. Second integral is difference between two paraboloids, but, they should take it over $D_2 \setminus D_1$.

Answer 3

$$\int\limits_{r=0}^3 2 \pi r (( 2 + 2 r^2/3) - \max[0,(r^2 - 1)])\ dr = 13 \pi$$