Log in ∑⨜| 10 years agoPosted 10 years ago. Direct link to ∑⨜|'s post “Why can't you just measur...” Why can't you just measure out each disc? • (0 votes) Stanley 10 years agoPosted 10 years ago. Direct link to Stanley's post “You have infinite disks, ...” You have infinite disks, and therefore, it would take a very, very long time... (28 votes) Strings to Eternity 11 years agoPosted 11 years ago. Direct link to Strings to Eternity's post “i do not understand how ...” i do not understand how does the definite integral comes to know of the exact variation of radius and how to add em up , • (7 votes) Rodrigo Campos 11 years agoPosted 11 years ago. Direct link to Rodrigo Campos's post “in technical terms, the d...” in technical terms, the definite integral is the Summation of the terms of a function with the first term equal to f(a), the second to f(a+dx), third to f(a+2dx) ..., term j equal to f(a+n.dx)=f(b), when the dx approaches 0, or you could say, when n approaches plus infinity. Each term is multiplied by dx, if you draw that you'll see that this Sum will give you the sum of many areas of rectangles. When the number of rectangles approaches inf. you will have a definite integral. Fortunately, someone proved that all of that I said equals to F(b) - F(a), when F(x) is the antiderivative of f(x) The Last Guy 11 years agoPosted 11 years ago. Direct link to The Last Guy's post “When finding the volume o...” When finding the volume of the "disk", why do you use the formula for the area of a circle, and not the volume of a cylinder? • (7 votes) Dan Surerus 11 years agoPosted 11 years ago. Direct link to Dan Surerus's post “Take a look at the Rieman...” Take a look at the Riemann sum that is a step before the definite integral. You can see there are physical cylinders that make up the volume for n= positive real number. You can use this formula as a better and better approximation for the volume as n becomes larger and larger. (5 votes) TQ 7 years agoPosted 7 years ago. Direct link to TQ's post “Can you use this for the ...” Can you use this for the y-axis as well? • (3 votes) Travis Bartholome 7 years agoPosted 7 years ago. Direct link to Travis Bartholome's post “Yes, you can use any of t...” Yes, you can use any of the methods around either axis. (That is, disks, shells, and washers.) You just have to keep in mind that your variable of integration has changed; so for example, if you're integrating dy and rotating around the x-axis, you'd have to use shells rather than using disks like you would if you were integrating dx. Hope that makes sense. (5 votes) Rehuel J. Galzote 10 years agoPosted 10 years ago. Direct link to Rehuel J. Galzote's post “What if the x-axis of rev...” What if the x-axis of revolution is moved? How would you deal with it? Any example videos please. Thank you. • (2 votes) nick.embrey 10 years agoPosted 10 years ago. Direct link to nick.embrey's post “You would still need to f...” You would still need to figure out the radius of the disks. The radius is the distance between TWO lines: the function f(x) and the axis of rotation. So if you were rotating around the line y=-2 (which is just shifting down two units from how it is now), then the radius of each disk would be x^2 + 2. (7 votes) Ionuţ Ştefan 11 years agoPosted 11 years ago. Direct link to Ionuţ Ştefan's post “So at 1:14 is the bottom ...” So at • (4 votes) jimstanley49 11 years agoPosted 11 years ago. Direct link to jimstanley49's post “The radii _are_ different...” The radii are different, but as the height becomes infinitely small, so does the difference in radius. It's much like the "problem" with Riemann sums, where the edges of the rectangles stick out to one side or the other of the function graph. As the rectangles get narrower and narrower, the error disappears. (3 votes) Gustavo Sáez 6 years agoPosted 6 years ago. Direct link to Gustavo Sáez's post “In the examples shown thu...” In the examples shown thus far f(x) is greater or equal to 0, but what of we had a cubic that dipped below the x-axis, would we still be able to use that formula? • (4 votes) Bryan 4 years agoPosted 4 years ago. Direct link to Bryan's post “Since you square r in the...” Since you square r in the formula, it doesn't matter if the function dips below. So yes! (2 votes) Soumya Sambeet Mohapatra 5 years agoPosted 5 years ago. Direct link to Soumya Sambeet Mohapatra's post “Hi Sal. A small question....” Hi Sal. A small question. Is there way to calculate the equation of the figure when y=x^2 is rotated about the x-axis? If there is a video could you please share it? • (3 votes) Kaleb Sagehorn 8 years agoPosted 8 years ago. Direct link to Kaleb Sagehorn's post “Say that the axis that we...” Say that the axis that we revolve our function around is also revolving around another axis. How would one calculate that? • (2 votes) Timothy S. Moran 6 years agoPosted 6 years ago. Direct link to Timothy S. Moran's post “Greetings!Actually, it ...” Greetings! Actually, it could not be an ellipse. You are rotating your y or (f(x)) value around the x axis. The absolute value of f(x) never changes, so the radius is constant as it rotates around the x axis. Thanks! (3 votes) Andrea Menozzi a year agoPosted a year ago. Direct link to Andrea Menozzi's post “if i were using the formu...” if i were using the formula for circumference instead of the area , would i find the surface area of the shape (without the end caps) correct? • (1 vote) Venkata a year agoPosted a year ago. Direct link to Venkata's post “Correct. Using the circum...” Correct. Using the circumference gives you the surface area. (3 votes)Want to join the conversation?
I think your question involves the limiting case that defines the definite integral as you let n go to infinity. The width of each cross sectional area "shell, washer...any shape" becomes infinitely small (dx). Limit as delta x approaches zero and the number of cross sectional areas you are adding up becomes infinite.
(dy)
1:14
So yes, the limit makes it a cylinder.
