Find the area between the two spirals r =θ and r = 2θ for 0 ≤ θ ≤ 2Π (two pie). Show your work?
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This can be represented as the double integral:\n[0, 2π]∫ [θ, 2θ]∫ 1 dA\nMaking the proper substitution of variables:\n[0, 2π]∫ [θ, 2θ]∫ r dr dθ\nAnd integrating:\n[0, 2π]∫ (2θ)²/2 - θ²/2 dθ\n[0, 2π]∫ 3θ²/2 dθ\n(2π)³/2 - (0)³/2\n4π³\n\nSteve: Actually, I did subtract in step 4 - note that the minuend is (2θ)²/2, not 2θ²/2. There is an important difference: the two is also being squared.\n\nAs for your answer - it cannot possibly be correct. The square [(2π, 0), (2π, -π√2), ((2+√2)π, -π√2), ((2+√2)π, 0)] has area 2π², and is contained entirely within the spirals (indeed, it only occupies a portion of the end), thus the area between them is clearly greater than 2π². Seeing a π³ term might be surprising, but it is quite correct.
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Crazydeadmoth got the hard half of the answer, but missed the easy half.  \n\nThe obvious solution is when x = PI/2 or 90 degrees.\n\nThe complete solution between 0 and 2pi is .284 radians and pi/2 (1.571 radians).
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The surface area of a sphere is 4 pi r^2 or 12.57 r^2.\nThe area of a curved surface of a spherical segment of height h, and radius of sphere r is equal to  2 pi r h\nSo, the area a=2 pi r h
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The area of the segment is given by\n\nA = r^2[theta - sin(theta)]/2\n\nSo we need theta, which is the angle that contains the chord, and the radius of the circle.\n\nThe radius of the circle is given by\n\nr = (c^2 + 4h^2)/(8h)\n= (20^2 + 4(6)^2)/(8*6)\n= (400 + 144)/48\n= 11.3 m\n\ntheta is given by\n\ntheta = 2 arcsin(c/[2r])\n= 2 arcsin(20/[2(11.3)])\n= 2.17 radians\n\nSo\n\nA = r^2[theta - sin(theta)]/2\n= (11.3)^2[2.17 - sin(2.17)]/2\n= 85.92 m^2
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Area increases 4 times. Beacause Area of circle = PiR2.\n\nSo if R doubles then radius will be 2R.\n\nHence area will be Pi(2R)2. which results into\n\n4 Pi R2.
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This is not a rigorous way to do it, but it works all the same:\nYou know that 4=y^2+x^2 is the equation for a circle of radius 2 centered at the origin.  The line y=0 is just the x-axis.  So find the area of the top half:  A = ½πr² = ½π(4) = 2π\n\n*If you had actualy integrated it, it would have been ugly:\nint[√(4 - x²)*dx] from x=-2 to x=+2.
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I assume you want to find the area between y = 0, y = x^3, x = 1, and x = 2.\n\nSince you're trying to find the area under a curve between 2 points on the x axis, just take the integral of the function from x = 1 to x = 2.\nintegral of x^3 = 1/4 * x^4\nf(1) = 1/4 * 1^4 = 1/4\nf(2) = 1/4 * 2^4 = 16/4 = 4\n\nf(2) - f(1) = 3 3/4 (solution!)
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I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
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I knew it as I did the activity.Draw a circle an divide it into 8 equal parts.It should be equal.Then cut it and stick such that the tip of one cone is up and the other is down.See that the edges stick each other close together.Do it until you finish all the 8 parts.\nNow you needed some concentration.You can see that it forms a parallelogram.Now area of parallelogram=base*height and the circumference of circle=Pi*2*radius.The base of the diagram is half the circumference that is Pi*radius,and the height will be the radius.You will also notice that Area of parallelogram=Area of circle.Therefore,Base*height=Pi *Radius *radius.so, now you have got pi*radius square.\nHope you like my answer.
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Take the formula for the surface area(S) and solve for r\nS=4*pi*r^2\nr^2 = S/(4pi)\nr =squareroot(S/4pi) or (S/4pi)^(1/2)\n\nNow take the formula for the volume\n4/3*pi*r^3\nand substitute the formula for r into the equation\n4/3*pi*(squareroot(S/4pi))^3\n4/3*pi(S/4pi)^(3/2)
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triangle area = 1/2(height * width)\n\ncircle area = pi * radius^2
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The curved surface area is 2*pi*r*h\n\nor 2*3.142*27*12 sq cms\n\nand 2*3.142*17*35 sq mms\n\nyou can do the rest
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2 X pi X  r = cirumference\n\npi X r squared = area
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area of rhombus=(1/2)d1d2\n(1/2)*16*d2=120\n8d2=120\ndividing by 8\nd2=120/8=15\nthe other diagonal is 15'
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Area of a circle = pi x r^2 =3.1416 x 4^2=50.265 cm^2\nArea of a circle = pi x (d^2)/4 = 3.1416 x (8^2)/4 = 50.265 cm^2\n\nCircumference of a circle = pi x diameter = 3.1416 x 8 =25.133 cm\nCircumference of a circle = pi x 2 x radius = 3.1416 x 2 x 4 =25.133 cm
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Area = pi r^2\nd Area/ dr = 2 pi r\n\nd Area = 2 pi r dr\n\nGiven d Area = 2 dr,  so substitute to get 2 dr = 2 pi r dr\n\n1 = pi r\n\n1/pi = r
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Area of the circle = pi r^2\nArea of the square = x^2\n\nThese are equal so x^2 = pi r^2\nTake the sqrt of both sides:\nx = sqrt(pi) r\nx = 1.77245385 r\n\nBut the diameter is twice the radius\nd = 2 r\n\nAs you can see d > x  because 2 > 1.77245385\n\nSo the diameter of a circle (d) would be larger than the side (x) of square with the same area.\n\nThe answer is:  The diameter of the circle