Volume by Integration. ...And we're back! This is the first problem of 2022, and I'm starting the year off with a problem type that's been on my mind for quite some time--finding the volume of an object via integration. As I reflect on being enrolled in Calculus 2 yeeeaaaarrrsss ago, I can recall learning this topic. It has been such a long time since I've done these types of questions. I'm super rusty, but I want to do it.
You can use a graphing calculator or other software to assist you with obtaining a visual of the solid, however, it is not required. I'm a visual person, so you know I'll be making use of graphing tools. To solve this week's problem in completion, you need to recall the following math skills:
✔️ How to find the area of a circle
✔️ How to integrate a function
WMP! #69 wants us to...
Check back on Saturday, January 15th for the solution, which will be posted below ⬇️.
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✏️📓 Solution Time! 📓✏️
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Let's go...
After reading the problem at least twice, I needed a visual to help me along.
The grey shaded area is the bounded region. The space to the left of the y-axis shows the outline of where the solid ends up when the shaded region is rotated about the y-axis. The figure we get looks to me like a tornado, except it would be filled in to be a solid. 🤷🏿♀️ How would you describe the shape of this generated solid? Leave your response down below.
Now that we know what kind of shape we're dealing with, we need to find it's volume. The cool thing about this solid figure is that it's horizontal cross-section is a circle...and we know how to find the area of a circle. (I did my best to show the circle in orange above in the visual aid.) Well, to find the volume of the figure, we need to integrate the expression that represents the area of the circle.
How do I know that radius is a function of y? The axis of rotation is the y-axis, which would we would see as the center of the circle from an aerial view. The length of the radius is the distance between the y-axis and the function which is the edge of the circle from the aerial view. So, the function needs to be in terms of y. To get the function in terms of y, isolate the variable x.
Cool. Let's move forward with the integration. To do so, we need one more thing...limits of integration. When you look at our figure, you can see that the circles are growing in area from the bottom of the figure to the top. Each circle has a different area at a different value of y, so the limits of integration start from the "smallest" circle (when y = 0) to the "largest" one (when y = 4).There you go!
▪️ Have you every used calculus to find the volume of a solid before?
▪️ Let me know what you thought about this week's problem in the comments section.
Thank you for solving with me this week. 😊
Up next is WMP! #70.
Cheers!
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