Find the area, in square units, of the region determined by the intersections of y = x^1/2 and y = x^2  /512 ?
925750
Sqtr(x)=x^2/512, x1=0, x2=64; \nI1=Integral [for x=0 until 64] of sqrt(x)*dx = 2x*sqrt(x)/3 = 16*(64/3)\nI2=Integral [for x=0 until 64] of (x^2/512)*dx = x^3/(3*512) = 8*(64/3)\nI1-I2=8*(64/3) – my answer!
85251
The area under any function can be got by the integration from an initial point to a final point\n\nfor example y(x)=x^2 and we are intrested to solve the area between x=1 to x=3\n\nthe integration = x^3/3 from 1 to 3\nthe area = (3^3/3) -(1^3/3) = 9 - 1/3 = 8.666667
694690
determine the square of 6 - sqrt[(x-2)^2 + y^2]\n\n-->\n\n36 + [(x-2)^2 + y^2] - 12*sqrt[(x-2)^2 + y^2]\n= 40 + x^2 + 2x + y^2 - 12*sqrt[(x-2)^2 + y^2]
301287
I am guessing you mean the area between the line y = 1-2x and the parabola y = x^2 + 2x + 1.\n\nTo do this, it takes a little calculus.\n\nFirst you find the points of intersection between the parabola and the line by setting the equations equal to each other.\n\nx^2 + 2x + 1 = 1 - 2x\n\nWe get x^2 = 4x = 0\nor x(x + 4) = 0 so the points of intersection occur at x = 0 and x = 4.\n\nAt this point you need to integrate each equation and evaluate between -4 and 0.  Subtract the area of the parabola between -4 and 0 from the line between -4 and 0 and you get the answer.\n\nThe integration for the line gives x - x^2 between -4 and 0\nThe integration for the parabola gives 1/3x^3 + x^2 + x between 0 and -4\n\nThis results in 20 - (64/3 - 16 + 4)= 32/3 square units\n\nCheck my math, but I think it's okay
48612
Let r be the radius of the circle and s the length of the square's sides.\ncircle perimeter=2*pi*r, square perimeter=4*s\n2*pi*r=4*s => s=pi*r/2\ncircle area=pi*r^2, square area=s^2\ns^2=(pi*r/2)^2=pi^2*r^2/4\nBut pi<4 => pi^2*r^2/4<pi*r^2 => square area < circle area
856302
Surface Area = (2 •  p   • r²) + (2 • p  • r • height)\n\nWhere (2 • p • r² ) is the surface area of the "ends" and\n(2 • p • r • height) is the lateral area (the area of the "side").\n\nVolume   =   p • r² • height   =   ¼ • p • d² • height\n\np = Pi
404470
a) AX = 2.25 cm.\n\nb) area AXY upon area ABC = 0.5625\n\nc) area AXY upon area trapezium XYCB = 0.4375
952353
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
342990
Take the derivative of each equation. Subtract one from the other. That'll get you the area enclosed between the two.
783536
let the length and width be 'x' and 'y' reply.\ngiven xy=512\nso x=512/y\nthe perimeter=2x+y\n=2x+512/x\nfor optimising\ndP/dx=0\n2-512/x^2=0\nmultiplying by x^2\n2x^2-512=0\n2x^2=512\ndividing by 2\nx^2=256\nx=rt256=16\ny=512/x=512/16=32\nso the dimensions for min perimeter\n=16'*32'\nperimeter=2x+y=64'\ncost=64*2=$128
1048814
If the circles are concentric, just subtract the area of the smaller circle from the larger circle:\n\nArea of small:  6*6*pi\nArea of large:  9*9*pi\n\nDifference of areas = 81*pi   -    36*pi  =  45*pi = 45 * 3.14159  = \n\n141.37 square units
876382
Area = (1/2) Base x Height.\n\nBase is x, because you can choose any side as base.  All are equal.\n\nThe Height (h), half of Base (0.5x), and one Side(x) form a right angled triangle.\n\nh/x = sin 60 = sqrt (3)/2\n\nh = (sqrt(3)/2)x.\n\nArea = (1/2)x (sqrt(3)/2)x\n=sqrt(3)*x^2/4
959978
use pythagorean thereom a^2 + b^2 = c^2. (c being teh longest side or the hypotenuse) the two legs are a and b. so you have the sq rt of 169. (sqrt of 12^2 + 5^2) which is 13
920664
area a to b is 1/2..\n\nhow? assume a side of the inner square as 1 so that its area would be also 1, by pythagorean theorem, half of the side of square b is equal to squareroot of 1/2, so the side of b is 2*squareroot(1/2), that makes the area to be equal to 2, getting the ratio of areas, that is 1:2 n_n
193440
Let x be the side of a square\nThen, area of the square is, A=x^2\nThe rate of change of area w.r.t side is dA/dx =2x\nIn the problem another parameter is given, that is time.It is given in the problem that there is a constant change in the length of the side at the rate 0.4cm /sec.Here we have to notice that corresponding change in area is not uniform throughout.Therefore we can only find instantaneous change in area or accumulated area in the given periodor interval.That means change in area at a given instant of time or interval.\nFor example, let initial area be 0.Now, to find the area between the interval a and b we have to reintegrate. ie. integration of dA between the interval a and b =integration of 2x in the interval a & b.\nie. [A]a&b= [x^2]a&b=b^2-a^2 square units\nFor instance, change in area between the interval 9th and 10th seconds is =[x^2]a&b=b^2-a^2 square units\nGiven, 0.4units/sec,considering intial length and time as 0,we have, length of the side on the 9th second is a=9*0.4=3.6units and in the same way length on the 10th second is b= 4units\nAccordingly change in area(accumulated area) at the 10th second is =b^2-a^2\n=16-12.96=3.04 square units [this is an instataneous change]\nPlease note that, this change cannot be written as units^2 /sec
513696
The area equation of a circle is pi r^2. And, a circle has an arc measure of 360. Since the shaded region has that of 40, it would be 40/360 = 1/9 of a circle.\nA = 1/9 pi r^2 = 59\npi r^2 = 59 x 9\nr^2 = 59x9/pi\nr = sqrt 59 x 9/pi = 13! =)
937424
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