The Disk Method Examples | Calculus 2 - JK Math

[@CryptoManDrei]

Veryyyy underrateddd calculusss tutorrrr ❤❤❤❤❤❤

[@brodyeskelsen4389]

Excellent Videos on disc and washer methods! helped me understand thanks!

[@JKMath]

You're welcome! Glad the videos were able to help you :)

[@mandandikaliyes847]

Understood now

[@AndreaPinedaMazariegos]

In the last example, there are two regions \between the curves , yet we only solved for one. How come? Great vid btw!

[@JKMath]

Good question! One of boundaries for the enclosed region of interest that was given is x=0, which is the same as the y-axis. So when considering that line, as well as the other two curves, the region that is explicitly enclosed by them is region we work with in the video, as all three curves/lines make up some portion of the boundary of the region (and also remember we are told to look in the first quadrant only). While the region below y=2x and between the x-axis is also in the first quadrant, the x-axis is not mentioned as one of our boundaries (or y=0). So, considering the description in the problem, the region we want must be between the given curves and x=0, and be in the first quadrant, and the region that best fits that description is the one we work with. Hard to explain well with just words, but I hope this helps paints a clearer picture of why we worked with the region we did. Feel free to ask any further questions if this is still not clear.

[@jeane.gancia2520]

Hii i would just like to ask, in the last example why was the two integrals added and not subtracted? don't we usually subtract two integrals?

[@JKMath]

In the last example we needed to break up the region being revolved around the y-axis into two smaller regions. We needed to do this because the outer function (or radius) of the region changes, it is not the same for the entire region. From y=0 to y=2, y=2x is the outer function, but then from y=2 to y=3, y=3-x² is the outer function. I explain this in more detail in the video, around the 27-28 minute mark. So, since we break up the region into two parts and we are calculating two different volumes that make up the solid of revolution, we need to add them together to get the volume of the entire solid of revolution. I think what you are referring to with subtracting integrals is when you are calculating the area under a curve or when using the washer method. In each of those cases we are subtracting the area/volume formed by a lower function that we do not want to include, which is not what is happening in this video. Hope this helps!

[@ritvikindupuri2388]

on the last example isn't it easier to use the washer method as there are two functions

[@JKMath]

Not quite. The inner radius is still 0, so we do need to use the disk method. There are two functions yes, but both functions form the outer radius. The outer radius is one particular function from y=0 to y=2, but then changes from y=2 to y=3, which is why we need to split it up into two integrals.

[@ritvikindupuri2388]

@@JKMath I see what about using thf shell method

[@JKMath]

@@ritvikindupuri2388 The shell method would make that example easier, yes. When revolving around the y-axis for the shell method, you want to work in terms of x, so you would not need to switch the functions into terms of y like we do for the disk method. The height of the shells h(x) would be the top function minus the bottom function, and the radius of each shell would be x. This should result in the same answer if you integrate from x=0 to x=1. I don't show the shell method in this video because this is a video on how to use the disk method. I discuss the advantages of the shell method in its lesson video. Hope this helps!

[@ritvikindupuri2388]

@@JKMath thanks always wanted to learn how to find height using shell. I also was wondering if you do any tutoring for calculus 2,, you are a great teacher

[@ritvikindupuri2388]

@@JKMath I see also for the radius is it always just x or y, like when does it change