Finding Volume: Rotating A Polar Curve Around The Y-Axis

Hey everyone! Today, we're diving into a cool problem from the world of calculus: figuring out the volume of a solid. Specifically, we're looking at the solid S that's created when we take a region defined by a polar curve and spin it around the y-axis. This process, called a revolution, creates a 3D shape, and our goal is to calculate its volume. Let's break down the problem step by step, making sure everyone can follow along. It's like building with LEGOs, but instead of plastic bricks, we're using mathematical concepts! We are going to address the question, how to calculate the volume of the solid S generated when rotating the region bounded by the curve r = 1 + cos(θ) situated in the first quadrant of the XY plane around the Y-axis?

First things first, we need to understand what our region looks like. The equation r = 1 + cos(θ) describes a cardioid in polar coordinates. Now, a cardioid is a heart-shaped curve. But, we're not dealing with the whole heart here; we're focusing on the part of the cardioid that lives in the first quadrant of the XY plane. This is the area where both x and y are positive. Imagine a heart tilted on its side, and we're just looking at the upper right portion of it. That’s the region we’ll be rotating. When we rotate this region around the y-axis, we're going to get a 3D solid that looks kind of like a rounded, lopsided vase. To calculate the volume, we're going to use a method that’s pretty common in calculus: the method of cylindrical shells. Think of it like this: we're going to slice our 3D vase into a bunch of thin, cylindrical shells, like the layers of an onion. Each shell has a certain radius, height, and thickness. By calculating the volume of each shell and adding them all up, we can approximate the total volume of the solid. The more shells we use, the more accurate our approximation becomes. That's the basic idea.

So, what does it really mean to use the method of cylindrical shells? Well, the volume (V) of a cylindrical shell is given by the formula V = 2πrh dr, where r is the radius of the shell, h is its height, and dr is its thickness (or width). In our case, since we're rotating around the y-axis, the radius r will be the x-coordinate of a point on the curve, which in polar coordinates is r cos(θ). The height h will be the y-coordinate of that point, which is r sin(θ). The thickness dr corresponds to an infinitesimal change in the angle θ, which can be seen as dθ. But, remember, we are in the first quadrant. So, the angle θ will range from 0 to π/2 (90 degrees). Now, we need to integrate this expression over the range of θ. The integral will give us the sum of all the tiny cylindrical shells that make up our solid. Let's put it all together. The volume of our solid S is given by the integral from 0 to π/2 of 2π times the radius, which is (r cos(θ)) times the height (r sin(θ)) dθ. But remember that r = 1 + cos(θ). Therefore, the volume becomes the integral from 0 to π/2 of 2π * (r * cos(θ)) * (r * sin(θ))* dθ = 2π ∫ from 0 to π/2 * (r *cos(θ)) * (r sin(θ)) dθ = 2π ∫ from 0 to π/2 * ((1+cos(θ))*cos(θ)) * ((1+cos(θ))*sin(θ)) dθ = 2π ∫ from 0 to π/2 * (1 + cos(θ))^2 cos(θ) sin(θ) dθ. That's the integral we need to solve.

Step-by-Step Calculation: Unveiling the Volume

Alright, guys, now comes the fun part: solving the integral! Don’t worry; we'll break it down step-by-step to make it super clear. Remember that our main goal is to solve the integral to find the volume of the solid that has been created. We've got the integral: 2π ∫ from 0 to π/2 * ((1 + cos(θ))^2 * cos(θ) * sin(θ)) dθ. To solve this integral, we'll use a technique called u-substitution, which is a handy trick in calculus. Let’s make a substitution: let u = 1 + cos(θ). Then, the derivative of u with respect to θ, du/dθ = –sin(θ). Rearranging this, we get du = -sin(θ)dθ. This is super useful because we have sin(θ)dθ in our integral! Also, we need to change our limits of integration, as our original integral used θ and we have changed to the variable u. When θ = 0, u = 1 + cos(0) = 1 + 1 = 2. When θ = π/2, u = 1 + cos(π/2) = 1 + 0 = 1. Now we can rewrite the integral in terms of u. The integral becomes 2π ∫ from 2 to 1 * (u^2 * cos(θ) * (-du))* = -2π ∫ from 2 to 1 * (u^2 * cos(θ)) du. Now we still have a cos(θ), which we need to get rid of. From u = 1 + cos(θ) we have cos(θ) = u - 1. So we can rewrite the integral again. It becomes -2π ∫ from 2 to 1 * (u^2 * (u - 1)) du. That’s much better because now our integral is only in terms of u. Simplify the integrand: (u^2 * (u - 1)) = u^3 - u^2. Now, our integral becomes -2π ∫ from 2 to 1 (u^3 - u^2) du. Next, we'll integrate each term separately. The integral of u^3 is u^4/4, and the integral of u^2 is u^3/3. So, the indefinite integral of (u^3 - u^2) is (u^4/4) - (u^3/3). We can plug in our limits of integration to calculate the definite integral: –2π [(u^4/4) – (u^3/3)] evaluated from 2 to 1 = -2π [((1^4/4) – (1^3/3)) - ((2^4/4) – (2^3/3))]. Now, let's plug in those limits! Evaluating at u = 1 we have (1/4) – (1/3) = –1/12. Evaluating at u = 2 we have (16/4) – (8/3) = 4 – (8/3) = 4/3. So we have -2π[(-1/12) - (4/3)] = -2π[(-1/12) - (16/12)] = -2π (-17/12) = 17π/6. Therefore, the volume of the solid is 17π/6 cubic units. That’s our answer!

This calculation process not only gives us the volume of the solid but also gives us practice in polar coordinates, integration techniques (like u-substitution), and the method of cylindrical shells. Remember, calculus is all about breaking down complex problems into smaller, manageable steps.

Practical Applications and Further Exploration

Okay, guys, so we've successfully calculated the volume! But where does this come into play in the real world? And what else can we explore related to this problem? Let's take a look. The principles we used here—calculus and volume calculations—are super important in a bunch of different fields. Engineers use these concepts to design everything from the shape of airplane wings to the capacity of storage tanks. Architects use them to calculate the space within buildings. Medical professionals use them in imaging to determine the size of tumors or the volume of organs. Even in the food industry, these principles help in designing packaging and figuring out how much of a product a container can hold.

Beyond just knowing how to do the math, understanding the why is also important. This problem isn’t just about the formula or getting a correct answer. It's about thinking critically and visualizing the problem in a new way. If you really want to expand on the topic, you can try: Explore different polar curves. The methods we used can be applied to many different shapes! Try calculating the volumes of solids formed by rotating other polar curves, like the rose curve or the limacon. Experiment with different axes of rotation. Instead of rotating around the y-axis, try rotating around the x-axis. Does this change the calculation process? How? Explore other methods of calculating volume, like the disk or washer method. In certain cases, these methods might be easier to use than cylindrical shells. You should investigate how the choice of method impacts the complexity of the calculation and the final answer. This helps solidify your understanding.

And most importantly: Practice! The more you practice, the more comfortable and confident you'll become. Each problem you solve is like leveling up in a game. You are building your knowledge base and expanding your problem-solving skills.

Key Takeaways and Wrapping Up

Alright, let’s wrap this up, guys! We started with a problem: finding the volume of a solid generated by rotating a polar curve around the y-axis. We used the method of cylindrical shells, which required us to set up an integral, use u-substitution, and evaluate definite integrals. We found that the volume of the solid generated by rotating the region bounded by r = 1 + cos(θ) in the first quadrant around the y-axis is 17π/6 cubic units. Remember that by breaking down the problem into smaller steps and using techniques like u-substitution, the most complex problems become solvable. By understanding the method of cylindrical shells, we can calculate the volume of a solid created by rotating a curve around an axis. We can apply this method to many other different shapes. This is just one example of the power and usefulness of calculus. Keep practicing, keep exploring, and keep having fun with math! You’ve got this!

In Summary:

• We calculated the volume of a solid of revolution. We used the method of cylindrical shells. The integral was solved by u-substitution.
• The final answer is 17π/6.
• We highlighted the method of cylindrical shells as a useful tool for this calculation. We explored different methods and looked at real-world applications of these concepts.

I hope you enjoyed this journey through calculus. Keep up the awesome work!