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 obtained by rotating the region bounded by the given curves about the specified l?
"Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 1/x^3, y = 0, x = 2, x = 4; about x = −1" Hi, could anybody give me some direction as how to solve this problem? I know that the integration is with respect to y, but I do not know what the inner and outer radii are. Would I have to put the 1/x^3 = y in terms of x? "Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves x=0, y=1, x=y^13 about the line y=1" Like the first one, this one, I assume, is with respect to x, so the integration seems more "normal." But how do I figure out the inner and outer radius? Are the points of intersection just found by plugging in the x lines into the y equation? How do the equations for the radius look if the solid is rotated at y=1? Many thanks, Cuanzo.
Find the volume of the solid obtained by rotating the region bounded by the curves y=cos(x) y=0 x=0 x=pi/2 abo?
Using the Washer Method: about y = 1 y = cos(x) A (x) = π ( outer radius )^2 - π ( inner radius )^2 A (x) = π ( 1 - 0 )^2 - π ( 1 - cos(x) )^2 A (x) = π ( 1 )^2 - π ( 1 - 2cos(x) + cos^2(x) ) A (x) = π ( 1 ) - π ( 1 - 2cos(x) + cos^2(x) ) A (x) = π ( 1 - 1 + 2cos(x) - cos^2(x) ) A (x) = π ( 2cos(x) - cos^2(x) ) π/2 ∫ π ( 2cos(x) - cos^2(x) ) dx 0 π/2 ∫ π ( 2cos(x) - (1/2) * ( 1 + cos(2x) ) ) dx 0 π/2 ∫ π ( 2cos(x) - (1/2) - (1/2) cos(2x) ) dx 0 . . . . . .. . .. . .. . .. . .. . .. . .. . .. . .π/2 π ( 2sin(x) - (1/2) x - (1/4) sin(2x) ) ] . . .. . .. . .. . .. . .. . .. . .. . .. . .. . .0 π ( 2 * ( sin(π/2) - sin(0) ) - (1/2) ( π/2 - 0 ) - (1/4) * ( sin(2 * π/2) - sin(2 * 0) ) ) ] π ( 2 * ( 1 - 0 ) - (1/2) (π/2) - (1/4) * ( sin(π) - sin(0) ) ) ] π ( 2 * 1 - (π/4) - (1/4) * ( 0 - 0) π ( 2 - (π/4) - 0 ) π ( 2 - (π/4) ) 2π - (π^2/4) ======== free to e-mail if have a question
Find the volume of the solid obtained by rotating the region bounded by the curves?
1) using the Disk Method: about y = 1 y = x^2/36 intersection: x^2/36 = 1 x^2 = 36 x = +/- 6 6 ∫ π ( 1 - x^2/36 )^2 dx = 16π/5 0 <---- x > 0 ---------------- 2) using the Disk Method: about x-axis y = 8x^2 1 ∫ π ( 8x^2 - 0 )^2 dx = 64π/5 0 <---- when y = 0 ---> x = 0 -------------- 3) Using the shell Method: height --------> 2 - √(x/2) radius --------> x intersections: 2 = √(x/2) 4 = x/2 8 = x 8 ∫ 2π * x * ( 2 - √(x/2) ) dx = 128π/5 0 -------------- 4) Using the Shell Method: 4 ∫ 2π * (4 - x) * ( √(x) - 0 ) = 256π/15 0 ------------ 5) using the Disk Method: about x-axis y = x^3 ; y = x intersection: x^3 = x x^3 - x = 0 x * (x^2 - 1) = 0 ----> x = 0, +/- 1 1 ∫ π ( x - x^3 )^2 dx = 8π/105 0 <---- x ≥ 0 =========== free to e-mail if have a question
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
Find the volume of the solid obtained by rotating the region bounded by curves?
Plot that -- y = tan²x tan x comes up from the origin starting with a slope of 1, and bends upward to ∞ at x = ½π. So it starts out close to y=x, before bending upward. And so tan²x starts out close to y=x², and so, looks a lot like that parabola, between x=0 and ¼π. So when you rotate about the x-axis, it turns into a round, pointy thing, with its point at the origin, pointing to the left. You need to get the volume from that point, rightward to the circular surface at x=¼π. To do this, you can set up little plates, each one going between x to x+dx, from x=0 to x=¼π. The radius of each plate is y = tan²x and so the volume is that of a cylinder, radius=y, height=dx: dV = π y²dx = π tan⁴x dx And the whole volume, V, is ¼π : : : ¼π ∫ dV = π ∫ tan⁴x dx x=0 : : : 0 : To do that integral, use trig identities and derivatives: tan²x = sec²x - 1 d(tan x)/dx = sec²x So tan⁴x = tan²x (sec²x - 1) = tan²x sec²x - tan²x ∫ tan⁴x dx = ∫ tan²x sec²x dx - ∫ tan²x dx which splits into these: ∫ tan²x sec²x dx = ∫ tan²x d(tan x) = ⅓tan³x + C₁ ∫ tan²x dx = ∫ sec²x dx - ∫ dx = tan x - x + C₂ so these recombine into: ∫ tan⁴x dx = ⅓tan³x + tan x - x + C making the definite integral: ¼π : : : : : : : : : : : : : : : : : : : : : : : : :| ¼π π ∫ tan⁴x dx = π(⅓tan³x + tan x - x) | = ⅓ + 1 - ¼π = 4/3 - ¼π ≈ 0.54793517... : 0 : : : : : : : : : : : : : : : : : : : : : : : : : | 0
How can I find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = sqrt (x − 1), y = 0, x = 4; about the x-axis?
It’s killing me to answer this without a blackboard to do a good explanation, but I’ll do the best I can without a good math editor on hand. You will be finding the volume using the disc method of integration. Each disc has a radius of sqrt(x-1) and a height of dx, so the volume of the solid will be the infinite sum of the volumes of all of these discs as dx becomes infinitely small, which will be the integral of the function pi sqrt(x-1) squared from 1 to 4 . Hopefully you know how to integrate because I’m skipping those details to give the answer 4.5pi.