Setting up integral using cylindrical volume?! CALC 2 HELP!?
(1 pt) The region bounded by y=x^6 and y=sin(πx/2) is rotated about the line x=−2. Using cylindrical shells, set up an integral for the volume of the resulting solid. The function to be integrated is:________
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the give?
Using shell method: 27y = x^3 -----> x = 3 y^(1/3) height --------> 6 - 3 y^(1/3) radius --------> 8 - y limits: when x = 6 ----> y = 8 8 ∫ 2π * (8 - y) * ( 6 - 3 y^(1/3) ) dy 0 <--- y = 0 8 ∫ 2π * ( 48 - 24 y^(1/3) - 6y + 3y^(1/3 + 1) ) dy 0 8 ∫ 2π * ( 48 - 24 y^(1/3) - 6y + 3y^(4/3) ) dy 0 . . . . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .8 2π * ( 48y - 24 y^(1/3 + 1)/(1/3 + 1) - 3y^2 + 3y^(4/3 + 1)/(4/3 + 1) ) ] . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. .0 . . . . . .. . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .8 2π * ( 48y - 24 y^(4/3)/(4/3) - 3y^2 + 3y^(7/3)/(7/3) ) ] . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. .0 . . . . . . .. . .. . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .8 2π * ( 48y - 24 * (3/4) * y^(4/3) - 3y^2 + 3 * (3/7) y^(7/3) ] . . .. . .. .. . .. .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. .0 . . . . . . .. . .. . . .. . .. . .. . .. . .. . .. . .. . .. . .8 2π * ( 48y - 18 * y^(4/3) - 3y^2 + (9/7) y^(7/3) ] . . .. . .. .. . .. .. . .. . .. . .. . .. . .. . ... .. . .. .0 2π * ( 48 * (8 - 0) - 18 * ( 8^(4/3) - 0^(4/3) ) - 3 * ( 8^2 - 0^2 ) + (9/7) * ( 8^(7/3) - 0^(7/3) ) 2π * ( 48 * 8 - 18 * ( 16 - 0 ) - 3 * ( 64 - 0 ) + (9/7) * ( 128 - 0 ) ) 2π * ( 384 - 18 * 16 - 3 * 64 + (9/7) * 128 ) 2π * ( 384 - 288 - 192 + (1152/7) ) 2π * ( 480/7 ) ( 960π/7 ) ======== free to e-mail if have a question
Find the volume of the solid obtained by rotating the region bounded by the curve y=sin(10x^2) and the x-axis?
Cylindrical Shells Method V = 2π ∫(0 to √(π/10)) x sin(10x²) dx Let u = 10x² ==> when x = 0, u = 0 ==> when x = √(π/10), u = π du/dx = 20x dx = du/(20x) V = 2π ∫(0 to π) x sin(u) * du/(20x) = (π/10) ∫(0 to π) sinu du = π/10 * 2 = π/5 ≈ 0.6283185 unit³
Cylindrical shells and volume?
1. Thickness of each shell = dy Radius of each shell = y Lateral surface area of each shell = 2 pi xy = 2 pi y sqrt(y) Volume of each shell = 2 pi y sqrt(y) dy = 2 pi [y^(3/2)] dy. After integration you have (4/5) pi [y^(5/2)] dy to be evaluated at 0 and 4: Volume = (4/5) pi (32) = 256 pi/5 [ about 160 ] #2. Rotating the region around what axis? #3. Thickness of each shell = dx. Length of each shell = sin(pi*x/2) - x^6 Radius of each shell = x - 4 Volume of each shell = 2 pi (x-4) [sin(pi*x^2) - x^6] dx So you have integral from x=0 to x=1 of 2 pi (x-4) [sin(pi*x/2) - x^6] dx
How can I find the volume of revolution when rotating [math] f(x)= \sin{(x-1)}, [0,1+ \frac{\pi}{2}][/math] around the [math]y[/math]-axis?
How can I find the volume of revolution when rotating [math]f(x)=\sin(x−1),\ [0,1+\frac{\pi}{2}][/math] around the [math]y[/math]-axis?The area enclosed is between [math]y=0[/math], [math]y=1[/math], and [math]x=0[/math].Here are two views of the region—one showing a horizontal rectangle (which when rotated around the [math]y[/math]-axis produces a disk), and the other showing two vertical rectangles (which produce cylindrical shells when rotated around the [math]y[/math]-axis).For the disk approach, the (infinitesimal) volume of the disk would be [math]dV=\pi \left(1+\sin^{-1}y\right)^2 dy[/math].For the shells approach, we could consider separately the two “types” of shells (those on [math][0,1][/math] and those on [math][1,1+\pi/2][/math]). The first set of shells gives a cylinder with height 1 and radius 1, while the second requires an integral; the volume of each of the second set of shells is [math]dV=2\pi x(1-\sin(x-1))\,dx[/math].Alternatively, we can take the volume of a cylinder with height 1 and radius [math]1+\pi/2[/math] and then subtract the volume of the “missing” piece (the solid formed by rotating around the [math]y[/math]-axis the area under the curve from [math]1[/math] to [math]1+\pi/2[/math]), for which the volume of a shell is [math]dV=2\pi x\sin(x-1)\,dx[/math].Therefore, the total volume is represented by any one of these expressionsDisks: [math]\int_0^1\pi \left(1+\sin^{-1}y\right) ^2 dy[/math]Shells (ver. 1): [math]\pi + \int_1^{1+\pi/2} 2\pi x(1-\sin(x-1))\,dx[/math]Shells (ver. 2): [math]\pi(1+\pi/2)^2 - \int_1^{1+\pi/2} 2\pi x\sin(x-1)\,dx[/math]All three give the same value (as one would expect) — about 8.196.
How can I find the volumes of the solid obtained by rotating the region bounded by y = x^2, y = 8, x = 0, about x = 3?
There was an error in the earlier version of the solution with the upper bound of y value. Thx kostyantyn!Rotate the red strip of thickness △x about the line x = 3Unfold the stripBy the fundamental theorem of calculusHope there are no mistakes
Help with integrals and rotating volumes?
First, draw the xy-coordinate system. Second, draw the horizontal line y = 0 and the curve y = sin x but only where x is between 0 and pi. Notice that the region is like a bell curve. Third, set up the integral for the volume of the solid obtained by rotating the region about the x-axis. Volume of the region = integral of {Pi * [sin x]^2 *} dx from 0 to Pi Note: The principle behind the formula of volume of a solid of revolution using the disk method is that you keep accumulating the volume of tiny cylinders starting from the left end of the integral limit until you reach the right end of the integral limit. Remember the volume of a cylinder is Pi * r^2 * h.
Let S be the solid obtained by rotating the region bounded by the curves y=sin(x2) and y=0, with 0x about the?
Region R : {(x,y) | 0 ≤ x ≤ √π ; 0 ≤ y ≤ sin x² } ; V = 2π ∫(x = 0 to √π) x sin²x dx = ...( cos 2x = 1 - 2sin²x ==> sin²x = (1 - cos 2x)/2 ) ... = 2π ∫(x = 0 to √π) x (1 - cos 2x)/2 dx = = π ∫(x = 0 to √π) x - x cos 2x) dx = = πx²/2 | (x = 0 to √π) - π ∫(x = 0 to √π) x cos 2x dx = ... u = x ; du = dx ... dv = cos 2x dx ; v = 1/2 sin 2x .. = π²/2 - π/2 x sin 2x | (x = 0 to √π) + π/2 ∫(x = 0 to √π) sin 2x dx = = π²/2 - π√π /2 sin 2√π - π/4 cos 2x | (x = 0 to √π) = = π²/2 - π√π /2 sin 2√π - π/4 cos 2√π + π/4 = ... - π/4 cos 2√π = - π/4 ( 1 - 2 sin² √π ) = - π/4 + π/2 sin² √π ... = π²/2 - π√π /2 sin 2√π + π/2 sin² √π ≈ 7.53 http://www.wolframalpha.com/input/?i=2%C...
How can I calculate the volume obtained by rotating the region bounded by the curves y=sinx, y=0, π/2 The solution is simple enough, if you consider that the volume is equal to the summation of the volumes of the differential cylinders of height dx and radius equal to the value of the y function (in this case, said y function is sin(x) ) at each x point from x=PI/2 to x = PI.So, all you have to do is to solve the integral of the said differential cylinders:The volume of the cylinder is PI*r^2*hIn this case r is equal to sin(x), and h is equal to dx, so, the integrand is:PI*(sin(x))^2*dxThe rest of the problem is simple, since you can either check the solution in a book with a table of common integrals, or use the properties of trig functions to convert (sin(x))^2 to an equivalent function that is useful for integration, ofr search the solution of said integral with the web app Wolfram Alpha.Kind regards, GEN