Question

Set up the triple integral for the volume of the solid that is the common interior below the sphere x2+ y2+ z2= 8 and above the paraboloid z= 1/2 (x2+ y2)

Answer 1

Answer:

$\iiint_V dx\,dy\,dz = \int\limits_{-2}^{2} \int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int\limits_{\sqrt{8-x^2-y^2}}^{\frac{1}{2}(x^2+y^2) } dx\,dy\,dz$

Step-by-step explanation:

The common region enclosed above the paraboloid $z = \frac{1}{2} (x^2+y^2)$ and below the sphere $x^2 +y^2+z^2 = 8$ is the solid.

Let's find the limit bounds for $z$.

The region is bounded above by the sphere $x^2 +y^2+z^2 =8$ .

Isolate $z$ from the equation above.

$x^2 +y^2+z^2 = 8 \iff z^2 = 8 - x^2 +y^2 \implies z = \sqrt{8-x^2-y^2}$

On the other hand, the region is bounded below by the paraboloid  $z = \frac{1}{2} (x^2+y^2)$

Therefore, we obtain

$\sqrt{8-x^2-y^2} \leq z \leq \frac{1}{2} (x^2+y^2)$

Now, we need to find the intersection of the sphere and the paraboloid. To do so, we need to solve the following system of equations

$x^2 +y^2+z^2 = 8$

$z = \frac{1}{2} (x+y) \implies 2z = x^2 + y^2$.

Substitute  the second equation in the first equation. We obtain

$2z+z^2 = 8 \implies (z-2)(z+4) = 0 \implies z = 2 \quad \vee \quad z = -4$

Hence, the paraboloid and the sphere intersect when $z= 2$.

Substituting $2$ for $z$ in the equation above gives

$x^2+y^2 = 4$

which is a circle with a radius $2$.

Now, we can find the bond for $x$ and $y$.

For $y$, we obtain

$\sqrt{-4-x^2} \leq y \leq \sqrt{4-x^2}$.

For $x$, we have

$-2 \leq x \leq 2$.

Therefore, the needed triple integral is

$\iiint_V dx\,dy\,dz = \int\limits_{-2}^{2} \int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int\limits_{\sqrt{8-x^2-y^2}}^{\frac{1}{2}(x^2+y^2) } dx\,dy\,dz$