Calculus Finding Volume Questions revolving?
1) Look here son. You're doing it all wrong! At most two questions per post. when doing this kind of problems a rough sketch is always helpful; shell method ````````````````` given y = x³ height of shell = y - 1 length of shell = 2πx ∴ Area of each shell = (length)(height) = 2πx(y - 1) = 2πxy - 2πx thickness of each shell = Δx enclosed area start from where y = 1 intersect y = x³ and ends where x = 2 x³ = 1 x³ - 1 = 0 x = 1 Volume V = ∫(2πxy - 2πx )dx from x = 1 to x = 2 V = ∫(2πx⁴ - 2πx )dx from x = 1 to x = 2 V = 47π/5 cubic units ================ (≈ 29.53 cubic units ) 2) again a rough sketch will be helpful (make a rough sketch to see how the lines and curves relate to each other) shell method: `````````````````` given y = x³ height of shell = 2 - x length of shell = 2π(10 - y) ∴ Area of each shell = (length)(height) = 2π(10 - y)(2 - x) thickness of each shell = Δy Volume V = ∫ 2π(10 - y)(2 - x)dy from y = 1 to y = 10 x = ∛(y) V = ∫ 2π(10 - y)(2 - ∛y) dy from y = 1 to y = 10 V = π(1233 - 450 ∛10)/7 V ≈118.26 cubic units ================
Calculus: finding volumes by revolving graphs around axes?
Can someone help with these two questions? a) Revolve the graph of sinx on the interval [0,pi] around the line y=2 b) Revolve the graph of 2-sinx on the interval [0,pi] around the x-axis
Finding the volume of a solid by revolving the region. Calculus help, which method should I use?
PROBLEM 1 The function y=x is bounded below by y=0, which is the x-axis. Since we are revolving it about the x-axis, it is not necessary to use the shell or the washer method; it's most convenient to use the Disk Method. Using vertical slices, we divide the volume up into a series of disks, each with radius r = y = x, and width dx. Then, each disk has a volume dV: dV = πr^2 dx = πx^2 dx The volume V is given by integrating: V = ∫ dV = ∫ πx^2 dx = πx^3 / 3 Evaluate this integral over x = 3 to 4: = π(4^3) / 3 - π(3^3) / 3 = 37π / 3 PROBLEM 2 On the interval x = 0 to 1, the region is bounded from above by y = 3, and from below by y = 3x. Rotating this about the x-axis, we get a horizontal cylinder with a cone-shaped "hole" through the center, so we use the Washer Method. Using vertical slices, we divide the volume into a series of washers, each with outer radius R = 3, inner radius r = y = 3x, and width dx. Then, each washer has a volume dV: dV = πR^2 dx - πr^2 dx = π(R^2 - r^2) dx = π(3^2 - (3x)^2) dx = π(9 - 9x^2) dx = 9π(1 - x^2) dx Integrating: V = ∫ dV = ∫ 9π(1 - x^2) dx = 9π ∫ 1 - x^2 dx = 9π (x - x^3 / 3) Evaluating over x = 0 to 1, we get: 9π [ (1 - 1^3 / 3) - (0 - 0^3 / 3) ] = 9π (2/3) = 6π Hope that helps!
How do I find the volume of a graph revolved around a line that isn't parallel or perpendicular to the x-axis?
You would rotate the graph so the axis of rotation would be parallel to the x-axis, then use normal methods.
Math calculus volume of revoltuion revolvign around y axis using washer's method?
First we need to find where they cross so set the two functions equal to each other.. x^3 - 1=x - 1 x^3 - x = 0 x(x^2 - 1) = 0 x=0,1,-1 are the solutions Now the curves will cross at (0,-1) and (1,0) I am going to assume that it is this region between these two points that is going to be revolved about the y axis, it is a moon shaped region. I have an idea that this is the region that they wanted to revolve, if we used all three we would need to add this second region to the first as a separate problem. If you look at the graph it makes alot more sense!!!! Since we are revolving around the y axis the width of the washers are dy so all varibles must be in terms of y. The basic equation is... V = ⌠(a to b) π [radius]²dy = π⌠(-1 to 0)[(cuberoot(y+1))² - (y+1)²]dy=π{3/5[y+1]^(5/3)-1/3[y+1]^3 }(-1 to 0)=π[3/5-1/3] -π[0]=4π/15≈.8378 We could double this answer and get the other region also. Good luck
Calculus Help? Finding volume by rotating function around x-axis?
In these questions, you need good visualisation in 3D to see what intergral(s) you are doing. (by the way, your first question's equation is wrong. It needs to be y = x/4, so it will cross (6, 24).) First, you need to separate the integral into continuous volumes. The only place where this would apply is your last question, where the graphs bounding above meet at x = 2, and where they are bound below by x = 1 and x = 3. So you will need to integrate from 1 to 2 and again from 2 to 3 with the respective equations. Now comes the integral part. You will need to find the "bigger" integral, do that, and the smaller integral from it. Usually this means integrating the upper-binding graph and subtracting the lower-binding graph. I'll do q3 for you as an example: y = e^x is definitely lower than y = 3. This graph will intercept at x = ln(3). So first we find the cylindrical area from x = 0 to ln(3) (no integral required, just a cylinder), which is area*height = pi*radius(this is the y value)^2 * height(this is the x value). Imagine the cone lying sideways. Then we have 3^2*pi*ln(3) = 9ln(3)pi. Next, we follow standard procedures to work out the "hole" in the cylinder created by y = e^x. Now we are integrating from x = 0 to ln(3) of pi*integral(e^(2x))dx. I trust that you can do this part yourself judging from your question. And that's it! Subtract what you have here from 9ln(3)pi and you have your answer. (Edit: I'm more inclined to help people who are aiming for self-improvement rather than just copy-pasting questions without attempting them first - which is why I wrote an essay for you :P)
Calculus Integral Problem HELP revolving function about axis find Volume: set up integral?
in general you have a choice when doing these revolution problems to use disks / washers or shells. Disks -- works particularly well when revolving about the x axis. f(x) is the radius of the disk. dx is the thickness. The volume is π f(x)^2 dx washers -- It is the same as disks, but there is a hole in the disk. Find the volume of the disk subtract the volume of the hole. Shells -- works best when revolving around the y axis. We take a cylindar of height f(x) and raduis (x+dx) and subtract a cylindar of height f(x) and raduis f(x) leaving behind a shell of thickness dx. The volume of this shell = π f(x) (x+dx)^2 - π f(x) (x)^2 = 2 π f(x) x dx But you are not locked into shells when rotating around the y axis and disk when rotating around the x axis. It could turn out that you want to use disks when rotating around the y axis. In which case, your disks would be. π f^-1(y)^2 dy To the problems at hand: Let f(x) = sqrt(16 - x^2) Let R be the region between the graph of f and the x axis on the interval [0,4]. Set up an appropriate integral to find the volume V of the solid region obtained by revolving R about the x axis and then find V. V = π∫ f(x)^2 dx = π∫ sqrt (16 - x^2)^2 | x = [0,4] = π ∫16 - x^2 dx 2. Let f(x) = sec(5x) Let R be the region between the graph of f and the x axis on the interval [0,pi/20]. Set up an appropriate integral to find the volume V of the solid region obtained by revolving R about the x axis and then find the volume (exact value). again, use disks... V = π∫ f(x)^2 dx = π∫ sec^2 (5x) | x = [0,pi/20] = π (1/5) tan 5x | x = [0,pi/20] π/5
Why do we use integration to find out the volumes of a ball revolving around the x-axis for example?
Answer: Because it works better than reading tea leaves.Next question: why do you use a can opener to open a tin of spaghetti? Answer: Because it works better than hitting it with your forehead.===Volume is DEFINED by integration. Namely, the ordinary “volume” enclosed in a boundary is the integration of that region against the canonical 3-form of flat euclidean space.Other quantities, such as mass, moment, polar-moment, and the coordinates of the centroid area also defined by integration.For some special cases, we have formulae that calculate volumes, moments etc. But those formulae were derived using calculus techniques. Calculus is a critical foundation of [almost] all of modern and classical mathematics.
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.
In calculus, can you revolve a function around a non-linear axis? If so, how?
In calculus, can you revolve a function around a non-linear axis? If so, how?You can’t rotate a function. A function is a rule that, given an input, produces an output. How do you rotate a rule?You can, however, rotate the graph of a function about a line. Each point on the graph rotates to produce a circle that is centred on the nearest point on the line. The line connecting the centre to your point is at right angles to the line.If you want to rotate a graph about a curve, I presume the centre of each circle will be the nearest point on the curve. But you might run into problems. Perhaps there is more than one nearest point, or perhaps you will create a surface that intersects itself. You could also suggest that the radius (line) of the circle should be perpendicular to the curve, just as it is for rotating about a line. But there could be even more such points.So it can be done, perhaps in more than one way. But you could end up with a mess. For example, suppose you rotate a circle about a concentric circle. If the circle you rotate about is bigger than your graph you will get a torus. If it’s smaller, you will also get a torus, but, depending on the radius, it could self-intersect. If your graph intersects the circle the resulting surface could be quite interesting (maybe not in a good way).I wonder in what sense rotating the graph of a function is calculus? Certainly you can use calculus to find surface areas and volumes etc., but these might not be very well defined if the surface self-intersects. But your question just refers to the set of points that the rotation produces. Not calculus, I aver.Edit: others have given a different interpretation. When a point on the graph moves in a circle, perhaps all points should move in circles on parallel planes. This is ambiguous until you decide what plane you want to rotate parallel to. Having decided, you want to choose the nearest (or often the only) point on the curve in that plane.