Math / Calculus

Question8. Evaluate $\iiint_{G}\left(x^{2}+y^{2}\right)^{3 / 2} \mathrm{dV}$ , where G is the solid enclosed between the paraboloids $z=2-x^{2}-y^{2}$ and $z=x^{2}+y^{2}$ .

Studdy Solution

STEP 1

What is this asking? We need to find the triple integral of $(x^2 + y^2)^{3/2}$ over the region $G$ trapped between two paraboloids. Watch out! Don't mix up the paraboloids!
One opens upwards, and the other opens downwards.
Also, cylindrical coordinates are our friends here!

STEP 2

1. Visualize and Set Up
2. Integrate!
3. Solve and Celebrate

STEP 3

Alright, let's visualize this!
We've got two paraboloids.
One, $z = x^2 + y^2$ , opens upwards like a bowl.
The other, $z = 2 - x^2 - y^2$ , opens downwards like an upside-down bowl.
They intersect somewhere, creating our region $G$ .

STEP 4

Let's find where they intersect!
We set the equations equal to each other: $x^2 + y^2 = 2 - x^2 - y^2$ .
Simplifying, we get $2x^2 + 2y^2 = 2$ , which means $x^2 + y^2 = 1$ .
This is a circle of radius 1 in the xy-plane!

STEP 5

Now, cylindrical coordinates!
Remember, $x = r\cos\theta$ , $y = r\sin\theta$ , and $z = z$ .
So, $x^2 + y^2 = r^2$ .
Our integral becomes $\iiint_G (r^2)^{3/2} \,dV$ , which simplifies to $\iiint_G r^3 \,dV$ .

STEP 6

Our $dV$ in cylindrical coordinates is $r\,dz\,dr\,d\theta$ .
So, the integral becomes $\iiint_G r^4 \,dz\,dr\,d\theta$ .

STEP 7

The bounds! $z$ goes from the bottom paraboloid ( $z = r^2$ ) to the top paraboloid ( $z = 2 - r^2$ ). $r$ goes from the center of the circle ( $r = 0$ ) to the edge ( $r = 1$ ). $\theta$ goes all the way around the circle, from $0$ to $2\pi$ .

STEP 8

Let's integrate with respect to $z$ first: $\int_0^{2\pi} \int_0^1 \int_{r^2}^{2-r^2} r^4 \,dz\,dr\,d\theta$ .
This gives us $\int_0^{2\pi} \int_0^1 r^4(2 - r^2 - r^2) \,dr\,d\theta$ , which simplifies to $\int_0^{2\pi} \int_0^1 (2r^4 - 2r^6) \,dr\,d\theta$ .

STEP 9

Now, let's integrate with respect to $r$ : $\int_0^{2\pi} \left[\frac{2r^5}{5} - \frac{2r^7}{7}\right]_0^1 \,d\theta$ .
Plugging in our bounds, we get $\int_0^{2\pi} \left(\frac{2}{5} - \frac{2}{7}\right) \,d\theta$ .

STEP 10

Simplifying the fraction inside gives us $\frac{14 - 10}{35} = \frac{4}{35}$ .
So, our integral is now $\int_0^{2\pi} \frac{4}{35} \,d\theta$ .

STEP 11

Finally, integrating with respect to $\theta$ gives us $\left[\frac{4\theta}{35}\right]_0^{2\pi}$ , which equals $\frac{8\pi}{35}$ .

STEP 12

Our final answer is $\frac{8\pi}{35}$ !

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