How can I find the volume of the solid generated by revolving the region bounded by [math]y=4x-x^2[/math] & [math]y=x[/math] about the y-axis?
This problem is a little tougher than some because the outer radius is determined by different curves for different values of [math]y[/math].Here's a plot to show what I mean.We need to rotate about the y-axis so we'll integrate the difference between the square of the outer radius and the square of the inner radius. The inner radius is always given by the x-coordinate of the left half of the parabola. The outer radius is determined by the x-coordinate of the line for [math]y\in(0,3)[/math] and by the x-coordinate of the right branch of the parabola for [math]y\in(3,4)[/math].Solving for the x-coordinate of the parabola as a function of y gives:[math]x^2-4x+4=4-y[/math][math](x-2)^2=4-y[/math][math]x=2\pm\sqrt{4-y}[/math]So the left side of the parabola is given by:[math]x=2-\sqrt{4-y}[/math]and the right branch is given by:[math]x=2+\sqrt{4-y}[/math]So here are the two integrals you must do:[math]V_1 = \pi\int_0^3 y^2 - (2-\sqrt{4-y})^2 \ \ dy [/math][math]V_2 =\pi\int_3^4 (2+\sqrt{4-y})^2 - (2-\sqrt{4-y})^2 \ \ dy [/math]Adding the results gives the answer.The first integral becomes:[math]V_1 = \pi\int_0^3 y^2 +y-8 + 4\sqrt{4 - y} \ \ dy [/math]I believe the result turns out to be [math]V_1 = \frac {49} 6 \pi[/math].The second integral becomes:[math]V_2 = \pi\int_3^4 8\sqrt{4-y} \ \ dy [/math].I believe the result turns out to be [math]V_2 = \frac{16}3 \pi[/math].Adding gives a total volume of [math]V=V_1+V_2 = \frac {27} 2 \pi \approx 42.41 [/math].I didn't check any of this, so hopefully someone will confirm that there are no errors in the calculations.
How would you find the volume of the solid generated by revolving around the line x=-4, the region bounded by that line and the parabola x=4+6y-2y^2?
volume of solid of revolution of x = 4 + 6*y(Note: this result is incorrect (see below))The limits of integration are found by calculating the points of intersection between the two curves (solving 4 + 6*y - 2*y^2 = -4 for y)You would be using disk integration, since integration is along the axis of rotation:Disc integrationFunction of y[edit]If the function to be revolved is a function of , the following integral will obtain the volume of the solid of revolution:where is the distance between the function and the axis of rotation. This works only if the axis of rotation is vertical (example: or some other constant).You will be calculating the distance of the curve x = 4 + 6*y - 2*xy^2 from x = -4 and using that for R(y).That distance is given by ABS(4 - 6*y - 2*y^2 - (-4)) = 8 - 6*y - 2*y^2. That is where the 8 term comes from in the Wolfram Alpha calculation.Note: I agree with Farough. The limits of integration should give a total length of 5 for the distance between the points of intersection:Intersection points of x = 4 + 6*y Now I'm confused as to why Wolfram Alpha gets the limits of integration tht are displayed above, giving a distance between the points of intersection of SQRT(17) < 5.It appears to give the correct answer if you move everything up four units, as Farough did (just adding 4 to the given function and rotating it about the line x = 0). I am not sure why it gets the wrong result when feeding it the problem as givenvolume of solid of revolution of x = 8 + 6*yIt also gives the correct answer with the unshifted functions if you force it to use the correct limits of integration as opposed to letting it determine them:Page on wolframalpha.com
Volume of revolution;y=x^2, y=4x-x^2 about the line y=6?
set them equal first as x^2 = 4x - (x^2) 2*(x^2) = 4x as x = 0 2x = 4 as x = 2 use the disk method as π * ∫_a^b R^2(x) - r^2(x) dx π * ∫_0^2 [(((x^2) - 6)^2) - ((4x - (x^2) - 6)^2)] dx simplify as π * ∫_0^2 [8*(x^3) - 40*(x^2) + 48x] dx π * [2*(x^4) - (40/3)*(x^3) + 24(x^2)] dx π/3 * [6*(x^4) - 40*(x^3) + 72(x^2)] 2π(x^2)/3 * [3*(x^2) - 20*x + 36] from F(2) - F(0) 2π(2^2)/3 * [3*(2^2) - 20*2 + 36] = 64π/3
Find the Volume of this Solid of Revolution. Please.?
1) working with the half above the y-axis, we have y = sqrt(4-x^2); by symmetry we just multiply by an additional factor of 2. Using the Shells method, dV = 2*pi*r*h*dr r = 3-x, h = y = sqrt(4-x^2) dr = dx V = 2*(2*pi* integral{-2 to 2, (3-x)*sqrt(4-x^2)dx }) =4*pi* [ 6*arcsin(x/2) - 1/6*sqrt(4-x^2)*(2x^2-9x-8) ] (-2 to 2) =4*pi*[6*(pi/2 +pi/2) - 1/6*(0 - 0) ] =24*pi^2. This can be checked, since it is the volume of a torus, and V = pi^2*r^2*R, with r = 2 and R = 3. 2) y = 6-2x-x^2 y = 6+x setting these two curves equal, x+6-6+2x+x^2 = 3x+x^2 = x(3+x) = 0 they intersect at x = 0,-3. Rotating about the x-axis, we use the Washer method dV = pi*(R^2-r^2)*dh dh = dx R = 6-2x-x^2 r = 6+x V = pi*integral{-3 to 0 [ (6-2x-x^2)^2-(6+x)^2 ]dx} =pi*integral{ -3 to 0 [ (x^4+4x^3-8x^2-24x+36) - (x^2+12x+36) ]dx} =pi*integral{ -3 to 0 [ x^4+4x^3-9x^2-36x]dx} =pi* [ x^5/5+x^4-3x^3-18x^2] (-3 to 0) =pi*(243/5-81-81+162) =pi*243/5 =152.68
What is the volume of the solid formed by rotating the region inside the first quadrant enclosed by y = x^2 and y = 5x about the x-axis?
We want the volume of the solid formed by rotating the region inside the first quadrant enclosed by the lines [math]y=x^2[/math] and [math]y=5x[/math] about the X-axis.We will first determine the points of intersection of these two lines so as to obtain the limits of integration.[math]y=x^2[/math] and [math]y=5x \qquad \Rightarrow \qquad x^2=5x\qquad \Rightarrow \qquad x(x-5)=0.[/math][math]\Rightarrow \qquad x=0[/math] or [math]x=5.[/math][math]\Rightarrow \qquad[/math] The points of intersection are [math](0,0)[/math] and [math](5,25).[/math]It can be seen that in the interval [math](0,5), x^2 < 5x.[/math]So, to get the volume of the required solid we have to rotate the area between the equations [math]y=5x[/math] and [math]y=x^2[/math] about the X-axis and integrate it from [math]x=0[/math] or [math]x=5.[/math]The area of the annulus formed by such a rotation with centre as any arbitrary point [math]x[/math] on the X-axis is[math]\qquad \pi\left(25x^2\right)-\pi\left(x^4\right)=\pi\left(25x^2-x^4\right).[/math]Consider a small slice of thickness [math]dx,[/math] which we obtain from the required solid between any arbitrary point x and [math]x+dx[/math] on the X-axis.The volume of this slice of infinitesimal thickness [math]dx[/math] is [math]dV = \pi\left(25x^2-x^4\right)\,dx.[/math][math]\Rightarrow \qquad[/math] The required volume is, [math]V = \int_0^5\pi\left(25x^2-x^4\right)\,dx = \pi \left[\frac{25x^3}{3}-\frac{x^5}{5}\right]_0^5[/math][math]\qquad = \pi \left[\left(\frac{25 \times 5^3}{3}-\frac{5^5}{5}\right)-\left(0\right)\right] = 416.67 \pi.[/math]
Find the volume generated by revolbing the area cut off from parabola y = 4x - x^2 by the x-axis?
This is essentially a washer. The outer radius, R, is the diffference between y=6 and the x axis (y=0) or 6. The inner radius r, is the difference between y=6 and the equation y=4x-x^2. The x axis and the function intersect at x=0 and x=4 The volume ,V is given by the integral V= pi int. (from 0 to 4) (R^2-r^2)dx or v= pi int. (from 0 to 4) ((6)^2-(6-(4x-x^2))^2)dx
Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines.?
a) Yes, the correct answer is 32π/3. b) One possible way is to shift the functions “down” by 8 (see the graph), and then revolve the new region about the x-axis, the result will be the same. New functions are y = x^2 - 8 y = -x^2 + 4x – 8 The volume of the solid is 2 ∫[π(x^2 - 8)^2 - π(-x^2 + 4x - 8)^2]dx = 32 π ≈ 100.53 0 -
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.
Calculus - Revolving an area about the line x = -1?
Really, both ways are almost the same. Cylindrical shells: radius x+1, volume of the shell [2 ln(x) - x/2]2π(x+1)dx => V = ∫(from x=2 to 8) [2 ln(x) - x/2]2π(x+1)dx = = 2π (80 ln(8) - 8 ln(2) - 141) = 124.471 Triple integral: When the figure rotates around the line x=-1, the a point of coordinate x₀ becomes the circle (x+1)² + z² = (x₀)². The solid whose volume is being calculated is defined as the region of the space which satisfies 3 < (√[(x+1)² + z²]) < 9 (√[(x+1)² + z²])/2