Find the volume of solid formed by revolving the region bounded by x=3(x^2) - 2x +1, y=3,about the line y=3?
This is a dumb problem to have to do using shells. It clearly calls for washers. If you MUST use shells, you have to do it in two steps because the specified curve defines both ends of the "height" of each shell. First you have to find the minimum of the curve. y' = 0 = 6x - 2 x = 2/3 WAIT. Before I go any further, please clarify: Did you mean y = 3x² - 2x + 1 (which I assumed), or x = 3y² - 2y + 1, which would be MUCH easier? EDIT: Thanks for the update. Using disks/washers, this is a fairly straightforward problem. First, let's find the limits of integration. That's where the curve intersects y = 3. 3 = 3x² - 2x + 1 0 = 3x² - 2x - 2 x = -0.5485837703548635, 1.2152504370215302 Then each disk will have a radius R = 3 - y = 3 - (3x² - 2x + 1) = 2 + 2x - 3x² (and no inner radius -- no hole, since we're bound by and rotating about y = 3), so V = π∫[-0.55, 1.22] (2 + 2x - 3x²)² dx = π∫[-0.55,1.22] (4 + 8x - 8x² - 12x³ + 9x^4)dx V = π(4x + 4x² - (8/3)x³ - 3x^4 + (9/5)x^5) |[-0.55, 1.22] V = π( 4.21 - (-0.91)) ≈ 5.12π ≈ 16 barring computational error. BTW: Thank you for the earlier BA. I added some comments to that question to help explain why there are two radii -- a symptom of all "washer" problems, where V = ∫ π(R² - r²) dx. (If there is only one radii, it's either a disk problem. where V = ∫ πR² dx, or a shell problem, where V = ∫ 2π R h dx.) Anyway, check out the comments; I hope they help. See link below.
Finding the volume of a solid based on the tan^3(x) function?
The integrals representing the volume of the solid obtained by rotating the region bounded by the curves y=\tan^3(x), y=1, and x=0 about the line y=1? y=tan^3(x), y=1 The point of intersection of y=tan^3(x) and y=1 is given by tan^3(x)=1 →x=π/4 Since rotation is about y=1 hence distances are to be measured from y=1 Since y=tan³x Hence distance between y= 1 and y= tan³x is 1−tan³x Therefore the integrals representing the volume of the solid obtained by rotating the region bounded by the curves y=\tan^3(x), y=1, and x=0 about the line y=1 is given as =∫(from x=0 to x=π/4) πy²dx =integral (from x=0 to x=π/4)∫ π(1−tan³x)²dx Hence F.
Sketch the region bounded by the curve and find volume revolving y-axis? ?
Part of the problem seems to be that the correct answer isn't listed. Perhaps choice c) should be (12/5)*pi, which is the correct answer. To calculate the volume of revolution, you first have to find where the curves intersect. This can be done by plotting or by setting the equations equal to each other and solving for (x,y). When you do this, you'll see that the functions intersect at (0,0) and (1,8). Here's a plot to visualize. http://i411.photobucket.com/albums/pp194... Since you're taking the volume of revolution about the y-axis, it's easiest to take your two radii at x-values, so that the lengths of the rectangles in your sum lie horizontally, whereas the width of each rectangle is dy. Therefore, expressing the two equations in terms of x, we have: For y = 8*sqrt(x) ..=>.. x = (y^2)/64 For y = 8x^2 ..=>.. x = sqrt(y/8) Also, note that x = sqrt(y/8) is greater than x = (y^2)/64 within the concerned region. Further, in terms of the y-axis, the range of interest is 0 to 8. That pretty much sets it up. So if you think of a given rectanangle, it will be "(pi*r^2) dy" where the r is defined by your functions. Surface of Revolution = IntOf_(0 tp 8) pi*(sqrt(y/8))^2 - pi*((y^2)/64)^2 dy = IntOf_(0 tp 8) pi*(y/8) - pi*((y^4)/4096) dy = IntOf_(0 tp 8) (pi/8)y - (pi/4096)y^4 dy = (pi/16)y^2 - (pi/20480)y^5 |0to8 = (pi/16)64 - (pi/20480)32768 = pi(64/16 - 32768/20480) = pi(4 - 8/5) = pi(12/5) <-- ANSWER !! I did this all out in Maple, to confirm that I arrived at the right answer. http://i411.photobucket.com/albums/pp194...
Find the volume of the solid obtained by rotating the region bounded by the given curves?
Shell method: V = 2π ∫ r h dr {a,b} .......... limits in {} radius is (1+y) to rotate around y=-1 = 2π ∫(1+y)(2) dx {0,1/3} + 2π ∫(1+y)(1/y - 1) dx {1/3,1} = 4π [y + y²/2] {0,1/3} + 2π [ ln(y) - y²/2 ] {1/3,1} = 14π/9 + 2π [ ln(3) - 4/9 ] = 2π [ ln(3) + 1/3 ] ≈ 8.997 units³ Disc method: V = π ∫ f(x)² dx {a,b} rotate around y=-1 extends the inner and outer radii by 1 = π ∫ (1/x + 1)² - (1)² dx {1,3} .............. note: π [(Ro)² - (Ri)²] = π ∫ 1/x² + 2/x dx {1,3} = π [ -1/x + 2 ln(x) ] {1,3} = π [ -1/3 + 2 ln(3) + 1 ] = 2π [ ln(3) + 1/3 ] Answer: ≈ 8.997 units³
Find the volume of the solid generated by revolving the region bounded by y=sqrt x and the lines y = 2?
It depends. Do you want to use the shell method ("dx" slices in this case) or the washer method? By shells: Each shell has radius R = 4 - x and height h = 2 - y = 2 - √x, so V = ∫[a,b] 2πRh dx = 2π∫[0,4] (4-x)(2 - √x) dx = 2π∫[0,4] (8 - 4√x - 2x + x^(3/2)) dx V = 2π(8x - (8/3)x^(3/2) - x² + (2/5)x^(5/2)) |[0,4] = 2π(32 - 64/3 - 16 + 64/5) V = (2π/15)(240 - 320 + 192) = 224π/15 By washers: x = y², and each washer has outer radius R = 4 and inner radius r = 4 - x = 4 - y², so V = ∫[a,b] π(R² - r²) dy = π∫[0,2] (4² - (4 - y²)²) dy = π∫[0,2] (8y² - y^4) dy V = π((8/3)y³ - (1/5)y^5) |[0,2] = π(64/3 - 32/5) = (π/15)(320 - 96) = 224π/15
What is the volume generated by rotating the area bounded by y=e^(2x), the y axis and the line y=2 about the x axis?
Using the disc method:First we find the intersection:[math]2=e^{2x}[/math][math]\ln 2 = 2x[/math][math]x = \dfrac{\ln 2}{2}[/math]Now we integrate:[math]\displaystyle V=\pi \int_0^{\frac{\ln 2}{2}} 2^2 - \left(e^{2x}\right)^2 dx[/math][math]\displaystyle V=\pi \ln 4 - \pi \int_0^{\frac{\ln 2}{2}} e^{4x} dx[/math][math]\displaystyle V=\pi \ln 4 - \pi (1-\frac{1}{4})[/math][math]\displaystyle V=\pi(\ln 4 - \frac{3}{4}) [/math]I purposely did the integration fast because you surely need more practice in it, and if you turned in your assignment as I did it, you would probably not get full marks. If you can't figure out how I did the integration, leave a comment and I can give some hints.
What is the volume of solid generated by revolving the area bounded by ellipse about its major axis?
The volume of an ellipsoid is similar to the volume of a sphere. For a sphere…[math]V=4/3*pi*r^3[/math]For an ellipse the major and minor axes take the place of the radius. For rotation around the major axis, the minor cross section is still a circle, the minor axis (b) is in the place of the radius two times:[math]V=4/3*pi*ab^2[/math]
Find the volume of the solid generated by revolving triangular region bounded by the lines y = 2x, y = 0, and?
The line y=2x cuts the axes at x = 0. We need to integrate between x = 0 and x = 1. Volume of revolution: V = pi S(0,1) y^2 dx = pi S(0,1) 4x^2 dx => V = pi[ (4/3)x^3 ](0,1) => V = pi( (4/3)(1)^3 - (4/3)(0)^3 ) => V = pi( (4/3) - 0) => V = (4/3)pi units^3
Find the volume of the solid obtained by rotating the region bounded by the given curves about the?
USING A DISK: dV = πy²dx V = π⌠y²dx limit 1 -> 5 but y = √(x-1) V = π⌠(√(x-1))²dx limit 1 -> 5 V = π⌠(x-1)dx limit 1 -> 5 V = π[x²/2-x] limit 1 -> 5 V = π[(5²/2-5) - (1²/2-1)] V = π[(5²/2-5) - (1²/2-1)] V = π(7.5+0.5) V = 8π ans USING A WASHER: dV = 2πy(5-x)dy V = 2π ⌠y(5-x)dy limit 0 -> 2 but y = √(x-1); x = y²+1 V = 2π ⌠y(5-(y²+1))dy limit 0 -> 2 V = 2π ⌠(4y-y³))dy limit 0 -> 2 V = 2π[4y^2/2 - y^4/4] limit 0 -> 2 V = 2π[2y^2 - y^4/4] limit 0 -> 2 V = 2π[2*2^2 - 2^4/4] - 0 V = 2π[8 - 4] V = 8π ans