Math / Calculus

Question5. Find the integral of $f(x, y, z)=8 z^{3}$ over the region in space bounded below by the $x y$ -plane and bounded above by the hemisphere $z=\sqrt{4-x^{2}-y^{2}}$ .

Studdy Solution

STEP 1

What is this asking? We need to find the volume under a 3D surface, which is like finding the area under a curve but one dimension higher!
It's bounded by the floor and a dome-like shape. Watch out! The hemisphere equation is given in terms of $z$ , so we'll need to be careful with our integration bounds.
Also, spherical coordinates will be our friends here!

STEP 2

1. Set up the integral
2. Convert to spherical coordinates
3. Evaluate the integral

STEP 3

Alright, let's define our integral!
We're integrating the function $f(x, y, z) = 8z^3$ .
Since we're dealing with a volume, we'll have a triple integral: $\iiint 8z^3 \, dx \, dy \, dz$

STEP 4

Now, let's figure out the bounds.
The region is bounded below by the $xy$ -plane, which means $z$ starts at 0.
The top is a hemisphere $z = \sqrt{4 - x^2 - y^2}$ , so $z$ goes up to that value.
For $x$ and $y$ , they cover the circle $x^2 + y^2 = 4$ on the $xy$ -plane, which is a circle with a radius of 2.

STEP 5

Spherical coordinates make this much easier!
Remember, $x = \rho \sin\phi \cos\theta$ , $y = \rho \sin\phi \sin\theta$ , and $z = \rho \cos\phi$ .
Also, $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ .

STEP 6

Our function becomes $8z^3 = 8(\rho \cos\phi)^3 = 8\rho^3 \cos^3\phi$ .
Don't forget to multiply by the Jacobian, $\rho^2 \sin\phi$ , so our integrand becomes $8\rho^5 \cos^3\phi \sin\phi$ .

STEP 7

Now for the bounds. $\rho$ goes from 0 to 2 (the radius of the hemisphere). $\phi$ goes from 0 (straight up) to $\pi/2$ (the $xy$ -plane). $\theta$ goes all the way around, from 0 to $2\pi$ .

STEP 8

Let's write our new integral: $\int_0^{2\pi} \int_0^{\pi/2} \int_0^2 8\rho^5 \cos^3\phi \sin\phi \, d\rho \, d\phi \, d\theta$

STEP 9

First, we integrate with respect to $\rho$ : $\int_0^2 8\rho^5 \, d\rho = \frac{8\rho^6}{6} \Big|_0^2 = \frac{8 \cdot 2^6}{6} - 0 = \frac{512}{6} = \frac{256}{3}$

STEP 10

Next, we integrate with respect to $\phi$ : $\int_0^{\pi/2} \cos^3\phi \sin\phi \, d\phi$ Let $u = \cos\phi$ , so $du = -\sin\phi \, d\phi$ .
When $\phi = 0$ , $u = 1$ .
When $\phi = \pi/2$ , $u = 0$ .
So, our integral becomes: $-\int_1^0 u^3 \, du = \int_0^1 u^3 \, du = \frac{u^4}{4} \Big|_0^1 = \frac{1}{4} - 0 = \frac{1}{4}$

STEP 11

Finally, we integrate with respect to $\theta$ : $\int_0^{2\pi} d\theta = \theta \Big|_0^{2\pi} = 2\pi - 0 = 2\pi$

STEP 12

Putting it all together: $\frac{256}{3} \cdot \frac{1}{4} \cdot 2\pi = \frac{128\pi}{3}$

STEP 13

The integral of $8z^3$ over the given region is $\frac{128\pi}{3}$ .

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