How do I find the volume of the solid by rotating the region (calculus 2)?
Find the intersection of the curves to determine the bounds of the region. 5 = x + 4/x 5x = x^2 + 4 x^2 - 5x + 4 = 0 (x - 1) (x - 4) = 0 x = 1, 4 There are two ways to do this: the washer method or the shell method. I'm going to use the shell method. http://www.wolframalpha.com/input/?i=y%3... Here's a graph of the region you're rotating. Picture the solid as a set of infinitely thin cylindrical shells, one inside of the other. The height of each shell will be the distance between y = 5 and y = x + 4/x; that is, it will be 5 - x - 4/x. The radius of each shell will be the distance between x and -1, or x - -1 = x+1. Each shell will have lateral area 2pi*r*h. To find the volume, integrate the lateral area over the bounds of the region. That is, V = 2 pi * integral(5 - x - 4/x)(x+1) from 1 to 4. Edit: I think you are right that the washer method will not work, or at least would be difficult to use. Take a look at the graph: how would you calculate the inner and outer radius of the washer? Both the outer and inner edges of the solid are defined by different points of the same curve, rather than two different curves. Try to rearrange the curve into a function of y, and you get x^2 - yx + 4 = 0; x = (y±√(y^2 - 16))/2. I suppose you could say that the value with the + is the outer radius and the value with the - is the inner radius. Then you'd have to find the minimum of the curve to find the lower bound of your integral. I chose to use the shell method because, after a quick moment of consideration, I could intuitively see how to apply the shell method while I hadn't quite worked through the process of the washer method in this case. Now that I'm considering it more closely, it was definitely the right choice.
CALCULUS HELP!!! Find the volume of a solid using integrals.?
I'll do #1 First graph the functions and find the region. Rotate this region about the x-axis to get yourself a good picture. Then you draw an arbitrary cross section of the region which will be a circle. The area of a circle is pir^2 so to find the volume you need to do the integral from -1 to 2 (the points at which the functions intersect) of pi((x+3)^2) -pi((x^2)+1)^2. (the functions are the radii of the circles) pi is a constant which you can put in front of the integral. so you are left with pi times the integral of (x+3)^2 - (x^2+1)^2 After you find the integrals multiply by pi and simplify you get -pi/5(x^5)-pi/3(x^3)+3pi(x^2)+8pi(x) [-pi/5((2)^5)-pi/3((2)^3)+3pi((2)^2)+8... Your numerical answer will be: 117pi/5 or 73.513
Calculus question - Volume of a solid?
Okedoke. First of all, the radius of the circle forming the base is 7. No argument, right? Therefore, a cross-section perpendicular to the x-axis at x=0 would yield a triangle with base of size 2r = 14 across. Therefore the cone is 14 units high. The volume of a right cone = 1/3 * pi * r^2 * h = 1/3 * pi * 49 * 14 = 686pi/3 cubic units. The area of a vertical cross section at x = 5? Well, what we need to do is find the size of the "base" of the triangle, which will also be its height, then we'll prettymuch be finished. For simplicity, imagine just the first quadrant of the x-y plane with the circular curve described by the function y = √(49 - x^2) When x = 5, y = 2√6. THAT is the length from the x-axis to one side of the circle (half the base of the triangle in the cross section). Cutting all the way across the circle will be 2 times this, or 4√6, as the base of the triangle. This will also be the height. Therefore, Area = 1/2 * (4√6) * (4√6) = 48 square units.
How do I find the volume of the solid by rotating the region (calculus 2)?
In the washer method, we find the area of an infinitely thin cross section of the solid and then integrate to sum the total volume of all such cross-sections in the solid. This problem requires the same basic concept, but the cross section will be a square. The equation of the circle is x^2 + y^2 = 1. For any given point x, there are two y values: ±√(1 - x^2). The side of the cross section at that point will be the distance between the y values, or 2√(1 - x^2). So the area of the cross section is 4(1 - x^2) = 4 - 4x^2 Integrate the area from -1 to 1 (the x-boundaries of the circle): integral (4 - 4x^2) = 4x - 4x^3/3 Evaluate from -1 to 1 = 16/3
Calculus II Problem: Volume?
Help on this subject would be very much appreciated because I don't fully understand all the material about computing the volume of an object through calculus. So, here's the question: ___________________________________ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specific line. Sketch the region, the solid, and a typical disk or washer. ______________________________________... Values: x=2sqrt(y), x=0, x=9; about the y-axis ______________________________________... I don't expect anyone to be able to sketch the object on a coordinate plane, but I would really appreciate if someone could explain to me what exactly I need to do so that I can sketch the figure. Thanks in advance for any help given.
Calculus! please help! finding the volume of a shape using disk!?
y = 6 - 6x² y = 0 Finding the limits of integration: 6 - 6x² = 0 6(1 - x²) = 0 x² = 1 x = -1 and x = 1 Using the disk method, the volume is described by the following integral: V = π ∫₋₁¹ (6 - 6x² - 0)² dx Or, taking advantage of the curve's symmetry about the y-axis, you can simply double half of the integral, meaning: V = 2 ∙ π ∫₀¹ (6 - 6x²)² dx V = 2π ∙ 36 ∫₀¹ (1 - x²)² dx V = 72π ∫₀¹ (1 - 2x² + x⁴) dx V = 72π [x - 2/3 x³ + 1/5 x⁵]₀¹ V = 72π [(1 - 2/3 + 1/5) - (0 - 0 - 0)] V = 72π (8/15) V = 192π/5
Calc 2 question!!! Find the volume of the solid obtained from rotating the region enclosed by...?
volume = π integral limit to limit [ f(y)^2 ] dy x = 4 - y x = 16 - y^2 for y = -3 to 4, 16-y^2 > 4-y volume = π integral -3 to 4 [ (16 - y^2 - (4 - y))^2 ] dy π * integral -3 to 4 [ y^4 - 2y^3 - 23y^2 + 24y + 144 ] dy π * ( y^5/5 - y^4/2 - (23y^3)/3 + 12y^2 + 144y ) [ -3 to 4 ] y = 4 ... 5312 / 15 y = -3 ... - 2061 / 10 π * (5312 / 15 - (-2061 / 10)) = 16807 π / 30 cubic units about 1760 cubic units
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.
Triple integral to find the volume of the solid bounded by the parabolic cylinder y=3x^2?
z-bounds: z = 0 to z = 7. This region projects onto the xy-plane as the region bounded between y = 3x^2 and y = 5 ==> y = 3x^2 to y = 5 for x in [-√(5/3), √(5/3)]. So, the volume ∫∫∫ 1 dV equals ∫(x = -√(5/3) to √(5/3)) ∫(y = 3x^2 to 5) ∫(z = 0 to 7) 1 dz dy dx = ∫(x = -√(5/3) to √(5/3)) ∫(y = 3x^2 to 5) 7 dy dx = ∫(x = -√(5/3) to √(5/3)) 7(5 - 3x^2) dx = 2 ∫(x = 0 to √(5/3)) 7(5 - 3x^2) dx, since the integrand is even = 14(5x - x^3) {for {x = 0 to √(5/3)} = (140/3) √(5/3). I hope this helps!