Consider the solid obtained by rotating the region bounded by the given curves about the line x = 6.?
y = √x is a real-valued function on the interval [0, ∞), so we may consider only nonnegative values of x. Using the method of concentric shells, we may integrate over x from x = 0 to 6. We divide this interval [0, 6] into two subintervals [0, 1] (where √x ≥ x), and [1, 6] (where √x ≤ x). Over [0, 1], each cylindrical shell of thickness dx has height h = √x - x and radius r = 6 - x. Its volume V₁ is: V₁ = 2πrh dx = 2π(6 - x)(√x - x) dx Over [1, 6], the height of each shell is h = x - √x, so its volume V₂ is: V₂ = 2π(6 - x)(x - √x) dx So, the total volume V is: V = ( ∫ V₁ dx | [ x = 0 to 1 ] ) + ( ∫ V₂ dx | [ x = 1 to 6 ] ) = 2π [ (14 / 15) + (554 / 15) - (48 / 5)√6 ] = 2π [ (568 / 15) - (48 / 5)√6 ] ∎ ≈ 90.17
Consider the solid obtained by rotating the region bounded by the given curves about the line x = 3?
Use shells to do this but disks will work as well. dV = 2π rdr h where r = (3 - x) is the radius measured from x = 3 dr = -dx y² = x/3, so y = ±√(x/3) and h = √(x/3) - (-√(x/3)) = 2√(x/3) dV = 2π (3-x) (-dx) 2√(x/3) = -4π (3-x) √(x/3) dx Limits: From r = 0 to r = 3 or x = 3 to x = 0 V = -4π ∫[x=3 to x=0](3-x) √(x/3) dx ...= -4π /√3 [x=3 to x=0](3*2/3 x^(3/2) - 2/5 x^(5/2)) ...= -4π /√3 {0 - (2*3^(3/2) - 2/5 * 3^(5/2))} ...= 4π (2*3 - 2/5 * 9) V = 48π/5 <=== Ans Hope this helps
Consider the solid obtained by rotating the region bounded by the given curves about the line y = 2.?
The region bounded by the given curves exists only between x=0 and x=1. However, since you give additional bounds as x=1 to x=3, there is a sort of triangular region whose right boundary is x=3, left or upper boundary y = 2x, lower boundary y = 2 sqrt(x). Although you could use the method of "washers," I find it easier in this case to calculate the volume of the outer cone by the simple formula V = (1/3) pi r^2 h = (1/3) pi (6^2) (2) = 24 pi, and then use integration to find the volume of the quasi-cone generated by rotating y = 2 sqrt(x) about the line y=2. Inner quasi-cone: Thickness of each disk = dx Radius of each disk = 2 sqrt(x) - 2 Area of each disk = 4 pi (x - 2 sqrt(x) + 1) Integrate from x=1 to x=3 4 pi (x - 2 sqrt(x) + 1) dx After integration you have 4 pi [ (1/2)x^2 - (4/3)x^(3/2) + x ] to be evaluated at x=3 and 1 = 4 pi [ (9/2) - 4 sqrt(3) + 3 - (1/2) + (4/3) - 1 ] = 4 pi [ 22/3 - 4 sqrt(3) ] = about 5.09 To see that this makes sense, note that the inner volume is slightly larger than a cone of radius 2 sqrt(3) - 2 and height = 2; the conical volume would be (1/3)pi(1.464^2)(2) = about 4.5 Final answer: V = 24 pi - 4 pi [ 22/3 - 4 sqrt(3) ] = 4 pi [ 6 - 22/3 + 4 sqrt(3) ] = 4 pi [ 4 sqrt(3) - 4/3 ] = 16 pi [sqrt(3) - 1/3] = about 70.3
Consider the solid obtained by rotating the region bounded by the given curves about the line x = 2.?
Hi, I'm back in yahoo answers. First, draw the graph of the parabola. Find its vertex and its opening. From the graph, you see that vertex V(x, y) = V(0, 0) and it opens to the right. Also, when x = 2, y = 2, -2. Rotating area bounded by the curve and x = 2 about x = 2, we obtain a disk. Using the shell method: ∫_(-1)^1〖 π r^2 h〗 = π ∫_(-1)^1〖4 y^4 dx〗 = π[4/5(1)^5 - 4/5(-1)^5 ] =8/5 π cubic units
Consider the solid obtained by rotating the region bounded by the given curves about the line y = 6.?
Use "cylindrical shells". The height of each shell is 3 - 3e^(-x). The radius of each shell is 6 - x. The thickness of each shell is dx. The volume of each shell is 2 pi RH dx = 2 pi [(6-x)(3 - 3e^(-x))] dx = 2 pi [18 - 3x - 18e^(-x) + 3x e^(-x)] dx The first three terms are easy to integrate; the fourth one requires "integration by parts"; anyway the integral of 3x e^(-x) is -[3e^(-x)](x+1). The limits on the integral are x=0 to x=5, and after integration you have 2 pi [18x - (3/2)x^2 + 18e^(-x) - [3e^(-x)](x+1)] = 2 pi [90 - 75/2 + almost 0 - almost 0] - 2 pi (0-0+18-3) = 2 pi (about 75/2) = about 236. Use the small terms too, if you want better accurately. The ANSWER can also be estimated by a Theorem of Pappus: the volume is the product of (a) the area and (b) the distance through which the centroid must travel. The centroid will be around x=3, so its orbit around x=6 will have a length 18pi, and the area looks like about 12.3, so I'd expect the answer to be about (3)(12.3)(2 pi) = about 230. This is very good news, it means I didn't screw up the arithmetic.
Consider the solid obtained by rotating the region bounded by the given curves about the x-axis?
a million. discover integration limits 2. discover which functionality is larger on integration era (further faraway from axis of rotation) 3. discover dV 4. combine to hit upon integration limits, discover the place the two curves intersect we are able to do this by way of comparing the place they have equivalent cost (y1=y2) 3x^6=3x 3x^6-3x=0 3x(x^5-a million)=0 it is genuine if the two 3x=0 or x^5-a million=0 in first case x=0 (because 3 is in no way 0) in 2d case x=a million we are able to %. any factor on era (different than ends) and notice which functionality is larger seems y=3x is larger than y=3x^6 on era [0,a million]. (try with x=0.5 as an instance) the physique would be hollow with outer radius R=3x and inner radius r=3x^6 then area of such slice is pi(R^2-r^2) and dV=pi(R^2-r^2)dx V=int[0,a million] dV
There are multiple ways to solve this, but I will use the "washers" method. The solid produced when rotating this region about the line [math]y=7[/math], when cut perpendicular to the x-axis, has cross sections that look like "washers", or circles with a circle cut out in the middle. To find the volume of the solid, one approach we can use is to express the area of one of these cross sections in terms of [math]x[/math], and then integrate this function from [math]x=-1 [/math]to [math]x=1[/math], since these are the endpoints of the region that we are rotating. So, essentially what we are doing is adding together the areas of the vertically oriented washers by finding the area of one washer in terms of [math]x[/math] and integrating this function from -1 to 1. This will give us the volume of the solid. Enough explaining; let's get into the math :) To find the area of a washer cross section, we will find the area of the outer circle and subtract the area of the inner circle. Since the axis of rotation is above the region we are rotating, the inner circle will be the one from the function [math]y=1[/math] and the outer circle will be the one from the function [math]y=x^2[/math] because [math]1 \geq x^2 [/math]for [math]-1 \leq x \leq 1[/math]. The radius of the inner circle is the y-value of the axis of rotation minus the y-value of the function since the cross sections are vertical. So, the radius of the inner circle is [math]7-1 = 6[/math] and the area of this circle is [math]36π[/math]. The radius of the outer circle is equal to 7 minus the y-value (in terms of x) of the function [math]y=x^2[/math] because this is the distance from the axis of rotation to the edge of the circle. Therefore, the radius is [math]7-x^2[/math] and the area of the circle is [math]π(7-x^2)^2[/math]. So, the area of a washer cross section is the area of the large circle minus the area of the small circle. Since we have already determined that we are integrating this area function from [math]x = -1[/math] to [math]x = 1[/math], we have [math]\int_{-1}^{1} (π(7-x^2)^2 - 36π \ dx[/math]From this point, it is just a matter of solving this integral. Here is how I did it if you are unable to solve this (I simplified it a bit at the start):It's [math]\frac{256π}{15}[/math][math] [/math] by the way ;)
Using cylindrical shell method (with horizontal slices)[math]V = 2\pi \displaystyle\int_1^5 (y-1) (4-(y-3)^2)\, dy[/math][math]= 2\pi \displaystyle\int_1^5 (-y^3+7y^2–11y+5)\, dy[/math][math]= 2\pi \bigg[-\dfrac{1}{4}y^4+\dfrac{7}{3}y^3–\dfrac{11}{2}y^2+5y\bigg]_1^5[/math][math]= 2\pi \bigg[\bigg(-\dfrac{625}{4}+\dfrac{875}{3}-\dfrac{275}{2}+25\bigg)-\bigg(-\dfrac{1}{4}+\dfrac{7}{3}–\dfrac{11}{2}+5\bigg)\bigg][/math][math]= \dfrac{128\pi}{3}[/math]Using washer method (with vertical slices)[math]V = \pi \displaystyle\int_0^4 [((3+\sqrt{x})-1)^2 - ((3-\sqrt{x})-1)^2]\, dx[/math][math]= \pi \displaystyle\int_0^4 (8\sqrt{x})\, dx[/math][math]= \dfrac{16\pi}{3} \Big[x^\frac{3}{2} \Big]_0^4[/math][math]= \dfrac{16\pi}{3} (8–0)[/math][math]= \dfrac{128\pi}{3}[/math]
V=π∫y²dx=π[math][0→1][/math]∫(36x^4)[math]dx=(0,1)[π×36x^5/5]=36π/5[/math][math]=7.2π[/math]unit³