area of region bounded by curves y=6x-x^2 & y=x^2-2x?
407819
Find the intersection points for the two curves.\n6x-x^2 = x^2 - 2x\n0 = 8x - 2x^2\n0 = x(8-2x)\nx = 0, 4\n\nIntegrate both functions\nfor 1st func.-->  y = 3x^2 - (x^3)/3\nfor 2nd func.-->  y = (x^3)/3 - x^2\n\nEvaluate integrals from x=0 to x=4 (intersection points)\nfor 1st integral--> A1 = 80/3\nfor 2nd integral--> A2 = 16/3\n\nTo find area of region bounded by both:\nA = A1 - A2\nA = 80/3 - 16/3\nA = 64/3\n\nThere's your answer!
501300
Area = Pi x R squared\nArea = Pi x (x - 6) squared\nArea = Pi x (x^2 - 12x + 36)\nArea = Pix^2 -Pi12x + Pi36\nArea = 3.14x^2 - 37.68x + 113.04\n\n(Using the approximation 3.14 for Pi)\n\nOf course, it has been a while and I'm a little rusty...
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
936604
let the equal sides be x\n1/2x^2=area=20 given\nx^2=40\nx=rt40\nusing the pythagoras theorem\nc^2=a^2+b^2\nc=rt(a^2+b^2)\nthird side=rt(40+40)\n=rt80\n=4rt5 cm
174683
The simple way to find the area of any shape is to break it up in to trangles and squares.  \n\n\nOr use integration... but i do not think that is what you are looking for
140062
L x W = Area\n\nCheater
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
657525
If you have the equation of the curve then integrate it on the lower and upper limits of the variable. In some cases you can find devide the identical parts and find out area of a single identical part and multiply by the number of parts made (sine wave).\n\nIf shape is a simple square circle etc you can find area by simple formula.
1148527
You need calculus. You can find that the places where the two curves intersect are at x=pi/3 and at x=-pi/3. So using the formula \nArea = integral (r^2/2), the area is:\n4(pi/3)n^2 - 4*(3^(1/2))*n^2 - 16n^2 (pi/3 + (3^(1/2))/2)
522898
Imagine you unraveled the cylinder, the middle would turn into a rectangle, right? so to find the surface area, take the height of the cylinder and multiply it by the circumference of the circles at the bases, then add in the area of the circles. Circumference is 2(pi)r and area is (pi)r^2.
612346
Fact or out the x. This leaves 36x^2-84x+49 which is a perfect square (6x-7) so it facotrs to:\n\nx(6x-7)(6x-7)
847764
Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)
752287
I'd suggest that you draw a picture. The area 'under' the inverse function can also be seen as the area to the *left* of the graph of y=f(x), i.e. x=f^(-1)(y).\n\nThe area under y=f(x) consists of a rectangle bounded by y=2, y=0, x=7 and x=11, and the area under y=f(x) and above y=2 from x=7 to x=11. By subtraction, this second area is 6.787151.\n\nNow the area to the left of y=f(x) consists of a rectangle bounded by y=2, y=5, x=0 and x=7 and the second area above. This gives a total of 21+6.787151=27.787151. This is the answer you seek.
342990
Take the derivative of each equation. Subtract one from the other. That'll get you the area enclosed between the two.
472218
I'm assuming that A and B are on the circle and O is the center of the circle.  Since OB=6, AB=6 as well, since OB and AB are radii.\n\n1.  Area of a circle = pi*r^2\n= pi*(6)^2\n= 36pi\n\n2.  Area of a sector (pie wedge) is easiest if you consider that the pie wedge is a certain fraction of the whole circle.  In this case, your sector is 60 degrees, and the whole circle is 360 degrees.  So the sector is 1/6 of the total area, which we found in part 1.  So the area = 6pi.\n\n3.  The segment is trickier.  If you look at the triangle, the segment is the curved part - almost like the top of the ice cream cone, if you can follow my description.\n\nThe area of a segment is given by\n\nA = r^2[theta - sin(theta)]/2\n\nwhere theta is the angle in radians.  The angle we have is 60 degrees.  To convert to radians, multiply by pi/180.  So theta is pi/3.\n\nNow since we know that the triangle has a 60 degree angle, and that OA and OB are both 6, through some reasoning we can see that this is an equilateral triangle, and AB (the chord) is also 6.\n\nSo the area of the segment then is\n\nA = r^2[theta - sin(theta)]/2\n= (6)^2[pi/3 - sin(pi/3)]/2\n= 3.26\n\n4.  Circumference = 2*pi*r\n= 2*pi*6\n= 12pi\n\nI don't think that X served any purpose except to indicate the region you're supposed to be considering for this problem.
587075
Would be happy to help, but some information is missing.  I am assuming that you are asking about the region bounded by the following four equations:\n{2x + y < 25}  {x + y < ?}  {x < 12}  {x > 0}\n\nNeed to know what ? is in order to give an exact answer.
1033128
Rewrite the equation as y=9/x. If we take one point on that curve and rotate it around the axis, the circle generated has a circumference of 18π/x and an area of  81π/x² (2πr and πr², respectively). Each of these circles is as an infinitesimally thin slice of the entire solid of revolution, and so we can find the surface area and volume of it by integrating the circumference and area of these circles, respectively. Thus the surface area of the entire curve is (n→∞)lim (1, n)∫18π/x dx (that should be read as "the limit as in approaches infinity of the definite integral from 1 to n of 18 pi over x with respect to x"). The antiderivative of 18π/x is 18π ln (x) and thus the by the fundamental theorem of calculus, (n→∞)lim (1, n)∫18π/x dx= (n→∞)lim (18π ln (n) - 18π ln (1)). Since the limit of ln n as n→∞ is ∞, the surface area is infinite.\n\nRepeating for the volume: the volume is (n→∞)lim (1, n)∫81π/x² dx . Our antiderivative is then -81π/x, and so the final volume is (n→∞)lim -81π/n - (-81π/1). Since the limit of 1/n as n→∞ is zero, the first term in this expression is zero, thus the final answer is - (-81π/1) or simply 81π.