Volume Calculation: Hemisphere And Cylinder Intersection

Hey guys! Today, we're diving into a super cool problem that involves finding the volume of a solid. This solid is bounded by a hemisphere and a cylinder. It might sound a bit intimidating, but trust me, we'll break it down step by step. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what we're dealing with. We need to find the volume of the region that lies inside both the hemisphere defined by the equation $z = \sqrt{16 - x^2 - y^2}$ and the cylinder defined by $x^2 + y^2 - 4x = 0$. Visualizing this in 3D space can be tricky, but here’s the gist:

• The hemisphere is the top half of a sphere with a radius of 4, centered at the origin (0, 0, 0).
• The cylinder is a cylinder whose axis is parallel to the z-axis. To understand it better, we can rewrite the cylinder's equation by completing the square. This will give us a clearer picture of its geometry. Let's dive deeper into this cylindrical shape!

Rewriting the Cylinder Equation

Okay, so let's tackle the equation of the cylinder: $x^2 + y^2 - 4x = 0$. To get a better handle on this, we're going to complete the square. This technique will help us rewrite the equation in a more recognizable form, revealing the cylinder's center and radius. Here's how we do it:

1. Rearrange the terms: $(x^2 - 4x) + y^2 = 0$
2. Complete the square for the x terms: To complete the square for $x^2 - 4x$, we need to add and subtract $(\frac{-4}{2})^2 = 4$. So, we get $(x^2 - 4x + 4) + y^2 = 4$
3. Rewrite as squared terms: Now we can rewrite the x terms as a square: $(x - 2)^2 + y^2 = 4$

Now, doesn't that look much clearer? This equation represents a cylinder with a radius of 2, whose central axis is parallel to the z-axis and passes through the point (2, 0, 0) in the xy-plane. So, we've got a hemisphere sitting on top of the xy-plane and a cylinder cutting through it. Our mission is to find the volume of the chunk of the hemisphere that's inside the cylinder. Time to bring out the big guns: calculus!

Setting Up the Integral

To calculate the volume, we're going to use a triple integral. Triple integrals are perfect for finding volumes in three-dimensional space. The general idea is to integrate over the region of interest, which in our case is the intersection of the hemisphere and the cylinder. The function we'll be integrating is simply 1, because we're just trying to find the volume (think of it as adding up infinitesimally small cubes).

However, before we can set up the integral, we need to decide on a coordinate system. Given the shapes we're dealing with – a hemisphere and a cylinder – cylindrical coordinates seem like the natural choice. Why? Because both shapes have nice, clean representations in cylindrical coordinates. Remember, cylindrical coordinates are an extension of polar coordinates into 3D, where we have $(r, \theta, z)$ instead of $(x, y, z)$.

Transforming to Cylindrical Coordinates

Let's quickly recap the transformations from Cartesian (x, y, z) to cylindrical $(r, \theta, z)$ coordinates:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$
• $z = z$ (z stays the same)
• $x^2 + y^2 = r^2$

Now, let's rewrite our equations in cylindrical coordinates:

• Hemisphere: $z = \sqrt{16 - x^2 - y^2}$ becomes $z = \sqrt{16 - r^2}$. This tells us how high up we need to go for a given radius r.
• Cylinder: $x^2 + y^2 - 4x = 0$ becomes $r^2 - 4r \cos(\theta) = 0$. We can simplify this to $r = 4 \cos(\theta)$. This equation defines the bounds for our radius r in terms of the angle $\theta$.

Determining the Limits of Integration

This is a crucial step! We need to figure out the ranges for $r$, $\theta$, and $z$ that cover the region we're interested in. Let's tackle them one by one:

• z: The height z varies from 0 (the xy-plane) up to the hemisphere, so $0 \le z \le \sqrt{16 - r^2}$.
• r: The radius r varies from 0 up to the cylinder's boundary, so $0 \le r \le 4 \cos(\theta)$.
• $\theta$: The angle $\theta$ determines how far around the z-axis we go. Since the cylinder is symmetric about the x-axis, we only need to consider the range where $r = 4 \cos(\theta)$ traces out the full cylinder. This happens when $\theta$ goes from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

With these limits in hand, we can finally set up our triple integral! The volume V is given by:

$V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4\cos(\theta)} \int_{0}^{\sqrt{16-r^2}} r \, dz \, dr \, d\theta$

Notice the extra 'r' in the integral? That's the Jacobian determinant for the cylindrical coordinate transformation, and it's essential for getting the correct volume. Okay, the integral is set up – now for the fun part: solving it!

Evaluating the Integral

Alright, let's roll up our sleeves and evaluate this triple integral. We'll tackle it step by step, starting from the innermost integral and working our way outwards.

Step 1: Integrating with Respect to z

The innermost integral is with respect to z:

$\int_{0}^{\sqrt{16-r^2}} r \, dz = r \left[ z \right]_{0}^{\sqrt{16-r^2}} = r\sqrt{16 - r^2}$

So, we've integrated out z, and now we have a double integral:

$V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4\cos(\theta)} r\sqrt{16 - r^2} \, dr \, d\theta$

Step 2: Integrating with Respect to r

Next up is the integral with respect to r. This one looks a bit trickier, but we can handle it with a u-substitution. Let's set:

$u = 16 - r^2$

Then, the differential du is:

$du = -2r \, dr$

So, $r \, dr = -\frac{1}{2} du$. We also need to change our limits of integration for r:

• When $r = 0$, $u = 16 - 0^2 = 16$
• When $r = 4 \cos(\theta)$, $u = 16 - (4\cos(\theta))^2 = 16 - 16\cos^2(\theta) = 16(1 - \cos^2(\theta)) = 16\sin^2(\theta)$

Now we can rewrite the integral with respect to r in terms of u:

$\int_{0}^{4\cos(\theta)} r\sqrt{16 - r^2} \, dr = \int_{16}^{16\sin^2(\theta)} \sqrt{u} \left( -\frac{1}{2} \right) du = -\frac{1}{2} \int_{16}^{16\sin^2(\theta)} u^{\frac{1}{2}} du$

Let's evaluate this integral:

$-\frac{1}{2} \int_{16}^{16\sin^2(\theta)} u^{\frac{1}{2}} du = -\frac{1}{2} \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{16}^{16\sin^2(\theta)} = -\frac{1}{3} \left[ u^{\frac{3}{2}} \right]_{16}^{16\sin^2(\theta)}$

Plugging in the limits, we get:

$-\frac{1}{3} \left[ (16\sin^2(\theta))^{\frac{3}{2}} - 16^{\frac{3}{2}} \right] = -\frac{1}{3} \left[ 64 \vert\sin^3(\theta)\vert - 64 \right] = \frac{64}{3} \left[ 1 - \vert\sin^3(\theta)\vert \right]$

Note: The absolute value is required because $\sin(\theta)$ can be negative in the interval $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, but the result of the cubic root must be positive.

So, our integral is now:

$V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{64}{3} \left[ 1 - \vert\sin^3(\theta)\vert \right] d\theta$

Step 3: Integrating with Respect to $\theta$

Finally, we need to integrate with respect to $\theta$. This one looks a bit intimidating because of the absolute value and the cubic power of sine. However, we can use symmetry to simplify things. Since both $\sin^3(\theta)$ and the integration interval are symmetric about $\theta = 0$, we can rewrite the integral as:

$V = 2 \int_{0}^{\frac{\pi}{2}} \frac{64}{3} \left[ 1 - \sin^3(\theta) \right] d\theta = \frac{128}{3} \int_{0}^{\frac{\pi}{2}} \left[ 1 - \sin^3(\theta) \right] d\theta$

Now, let's break this into two integrals:

$V = \frac{128}{3} \left[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \int_{0}^{\frac{\pi}{2}} \sin^3(\theta) \, d\theta \right]$

The first integral is straightforward:

$\int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$

The second integral, $\int_{0}^{\frac{\pi}{2}} \sin^3(\theta) \, d\theta$, requires a bit of trigonometric manipulation. We can rewrite $\sin^3(\theta)$ as $\sin(\theta) \sin^2(\theta)$, and then use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$:

$\int_{0}^{\frac{\pi}{2}} \sin^3(\theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} \sin(\theta)(1 - \cos^2(\theta)) \, d\theta$

Now, we can use another u-substitution. Let's set:

$u = \cos(\theta)$

Then, $du = -\sin(\theta) \, d\theta$. We also need to change our limits of integration for $\theta$:

• When $\theta = 0$, $u = \cos(0) = 1$
• When $\theta = \frac{\pi}{2}$, $u = \cos(\frac{\pi}{2}) = 0$

So, our integral becomes:

$\int_{0}^{\frac{\pi}{2}} \sin(\theta)(1 - \cos^2(\theta)) \, d\theta = -\int_{1}^{0} (1 - u^2) \, du = \int_{0}^{1} (1 - u^2) \, du$

Evaluating this integral, we get:

$\int_{0}^{1} (1 - u^2) \, du = \left[ u - \frac{u^3}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$

Putting it all together, we have:

$V = \frac{128}{3} \left[ \frac{\pi}{2} - \frac{2}{3} \right] = \frac{128}{3} \left( \frac{3\pi - 4}{6} \right) = \frac{64(3\pi - 4)}{9}$

The Final Answer

Woohoo! We made it! The volume of the solid bounded by the hemisphere and the cylinder is:

$V = \frac{64(3\pi - 4)}{9}$ cubic units.

That was quite a journey, guys! We started by understanding the problem geometrically, then we transformed to cylindrical coordinates, set up a triple integral, and finally, carefully evaluated it step by step. This problem showcases the power of calculus in solving complex geometric problems. I hope you found this explanation helpful and maybe even a little fun! Keep exploring the fascinating world of math!