Calculus: Shell Method?
Cylindrical shells are a good method to find the volume of revolution. -take a function -rotate it around an axis -you need to get a volume How you can know when to use this is when you need to integrate along an axis that is PERPENDICULAR to the axis of revolution. Like let's say you are rotating around the y-axis, but you need to integrate dx. This is hard to "see" so try drawing a picture. I found this: http://www.stewartcalculus.com/data/CALC... It's a pdf, but it opened in my browser without downloading, and it has good pictures that make sense, I think. To understand how to use it, always first picture or draw the solid of revolution, and then draw a REPRESENTATIVE RECTANGLE. Then use the above picture to set up your integral. I'm sorry I know this is *really* hard to explain without being able to draw a picture. :( But try to check out the pdf above - it's not bad. Meanwhile, I'll dig around and see if I can find any more pictures online to link to. If I can find anything else, I'll add it under the references section...
Volume- Shell Method (f(x)=1/x) [Calculus]. Help?
Hello. We were told to find the volume of f(x)=1/x; on interval [1,4] using shell method. I thought it would be like any regular shell method problem, but it says that if you use shell method you need to divide the integral into two parts of different heights but same radius...why ? Is this a special case ?
Shell method calculus help?
Set up the integral for finding the volume by specified method for specified axis for the region bounded by y = 1 + x^2, y = 0, x = 0, x = 2 a. shell method around x axis b. disk method around y axis c. disk method around the line y = 4 d. method of choice around x =4 Be sure to show all work and explain. Thanks.
Using the shell method with vectors! calculus.?
It's easiest to look on the y-z plane to see the volume of revolution. It is the upper portion of the sine curve, from 0 to π, revolved around the z-axis. A differential area element is then dA = z dx. Revolving this around the z-axis gives the differential volume element, dV = 2πr*dA dV = 2π*y*(z dx) dV = 2π*y*sin(y) dx Then, integrating (I used integration by parts, u = y, dv = sin(y) dy) V = 2π{-y*cos(y) + sin(y)} evaluated between π and 0 V = 2π{ (-π*cos(π) + sin(π)) - (-0*cos(0) + sin(0)) } V = 2π² Then, to 3 decimal places V = 2(3.14159265)² V = 19.739 It is educational to use the Theorem of Pappus to check the result. This theorem tells us to take the area of the cross section and revolve it along a circle with the same radius as the centroid of the cross section to get the volume. What's the area under the sine curve from 0 to π? A = Integrate: sin(y) dy = -cos(y) = -cos(π) + cos(0) = -(-1) + 1 A = 2 What is the radius of the centroid of the sine curve from 0 to π? R = π/2 Then, by Pappus V = 2π*R*A = 2π(π/2)(2) = 2π² So, we get the same answer both ways!!
Using the shell method to calculate the volume of a solid. Calculus help?
It's going to be way easier to integrate with respect to x, since the axis of rotation is vertical, the shell method requires slicing vertically, so we're good. Note that the curves intersect at x = 0 and x = 2. V = 2π ∫ rh(x) dx from 0 to 2 First determine the radius. When x is 0 , the radius is 0 and when x is 2 the radius is 2, so the radius is just x. Now determine the height. Since 8 - x^2 is above x^2, subtract x^2 from 8 - x^2. h(x) = 8 - x^2 - x^2 = 8 - 2x^2. Now sub all of that in: V = 2π ∫ [x][8 - 2x^2] dx from 0 to 2 V = 2π ∫ [8x - 2x^3] dx from 0 to 2 V = 2π[4x^2 - (1/2)x^4] from 0 to 2 V = 2π[4(2)^2 - (1/2)(2)^4] <-- note that the lower limit of zero will be zero V = 16π Done!
Calculus - washer vs. shell method - Warning: Not easy?
I think that you meant discs and shells methods. Also I assume that the region below the curve is shaded. Shells method Vol = 2pi*INT xy dx = 2pi*INT(0, 2) x + (x^3)/4 dx = 2pi*[(x^2)/2 + (x^4)/4] (0, 2) = 2pi*[(2 + 1) - (0 + 0)] = 6pi Discs method. The open bowl shape that you have is a cup shape subtracted from a cylinder. The cylinder has radius 2 and height 2 so volume 8pi. The cup shape is a y axis disc integral. x = 0, y = 1 and x = 2, y = 2 so new limits are 1, 2 Vol = pi*INT x^2 dy = pi*INT(1, 2) 4(y - 1) dy = pi*[2y^2 - 4y] (1,2) = pi*[(8 - 8) - (2 - 4)] = 2pi When this is subtracted from the 8pi of the cylinder you again get 6pi.
Determine the volume of a torus using the shell method [calculus]?
The trick of these integrals is usually to substitute for a cosine but you must have a term which is always less than 1 in absolute value. here you can rewrite the function as x*r*sqrt(1-((x-a)/r)^2)dx if (x-a)/r = sinT then dx/r = cosTdT or dx = rcosTdT and x = rsinT + a. for x = b+a. T = Arcsin(b/r) and for x = a-b, T = Arcsin((b-2a)/r) You get: (rsinT + a)*r*sqrt(1-sin^2T)*r*cosTdT = (rsintT + a)*r^2*cos^2TdT Hopefully you can integrate that one. I am not sure what your term b is though. I am assuming that an integration in parts should work.