area of the region bounded by the curves y=x^2 & y=-2x^2+27?
457571
the answer a:108\nbecause the intersection between the two curve ;-3and3\n[  x^2=-2x^2+27   the value of x are 3 and -3  ]\nyou will integrate from -3 to 3 (-2x^2 +27 - (x^2)) dx\n[-x^3+27x] from -3 to 3 \nsubtitute you will get 108
928305
The volume V=abc\nSurfce area S=2(ab+bc+ca)\nTherefore,2/S(1/a+1/b+1/c)\n=2/2(ab+bc+ca )](ab+bc+ca)/abc\n=1/abc=1/V proved
288086
I assume this is a question from a calculus class.  The only way I know of solving this is using derivatives.\n\nsurface area = 2Pi*r^2 + 2Pi*r*h\n\nV = Pi*(r^2)*h\nV = 400Pi\nThus, 400Pi = Pi*(r^2)*h  Solve this equation for h by dividing both sides by Pi*r^2\n400PI/Pi*r^2 = h  Simplifying will give you\nh = 400/r^2\nNow replace h with 400/r^2 in the equation \nsurface area = 2Pi*r^2 + 2Pi*r*h\nLet S stand for surface area\nS = 2Pi*r^2 + 2Pi*r*(400/r^2)  Simplify\nS = 2Pi*r^2 + 800Pi/r\nNow take the derivative.\nS' = 4Pi*r - 800Pi/r^2\nSet this equal to zero and solve for r.\nI get that r = the cube root of 200 which is 5.84803547643\nNow you should test to see whether this is a max or min.  It should be a min; although, I did not check it.\nAfter you check it plug this value of r = 5.84803547643 into the equation h = 400/r^2\nh = 400/(5.84803547643)^2 \n= 11.6960709529\nThus,\nr = 5.8480354764cm and h = 11.6960709529cm
1019108
Do you have limits between which to calculate the area? If not you can solve for the values where f(x) = g(x)\nyou should get two values since g(x) is a second degree function.\nThe 2 values are your limits for integration.\nIntegrate each function with respect to dx between the limits and take the difference of the two. Since area is always positive the value should be the absolute value of the difference.\n\nI think that is enough hint. I won't do it for you. But here is what you should start with:\n\nsolve for x where f(x) = g(x), the two values for x are the two limits for your integration, for example x1 and x2.\nthen integrate\n  x1\n| ∫( f(x) - g(x) ) dx |\n  x2
985185
X                      Y1                        Y2\n-2                     -4                          7\n-1                     -1                          6\n0                       2                           5\n2                       8                           3\n5                       17                         0\n\nX is your domain and Y1 is the range for y=3x+2 and Y2 is the range for y=-x+5.\n\nHope this is what you are looking for.
103215
With this given data, it is not possible\n\nalthough the area is totalling to be 70 for both cases (7*10 and two pieces of 8*8 + 1*6) but with one division only, this is not possible\n\nPLEASE DO LET ME KNOW IF THERE IS SOME SOLUTION (i found it very interesting)
755101
Well area is equal to Radius squared times 2 pie(or22/7)\nSo you would have 35^2=1225 and then 1225*2*(22/7)=7700 inches for your area\n\nhope that helped
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
863802
area=(1/2)d1*d2\n=(1/2)*28*13\n=14*13\n=182 sft
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)
668045
The only finite region bounded by those graphs is between the only two points at which they intersect, (0, 0) and (1, 1). f(x, y) = x + y is a plane x + y - z = 0 that passes through the origin. Note that in R, y=x^2 is below y=x with respect to the x-axis.\nNow that we have our bearings, we simply decide the boundary of our region and integrate over it. We are integrating from x=0 to x=1 and from y=x^2 to y=x the form x + y dy dx.\nIntegrating with respect to y first, we get the form xy + (y^2)/2 evaluated from y=x^2 to y=x which is (3/2)x^2 - x^3 - (x^4)/2. Integrating this form with respect to x gets us (1/2)x^3 - (x^4)/4 - (x^5)/10 evaluated from x=0 to x=1, which is just 1/2 - 1/4 - 1/10 = 3/20.
921006
Area = πab where a and b are the semi-major and semi-minor axis length
265577
LENGTH X WIDTH X HEIGHT DIVIDED BY 27
71760
area := Πr2\ncircumference := Πd  ,  r=radius, d=diameter, Π=3.14128....
76067
The answer is 13.5 in.\n\nFirst we have that the total area is 178 in^2, and we have the lenght and width, so we just need to write the formula for the area of each side and sum them. We would have an unknown variable: height.\n\n178 = (2*4) + (2*4) + (4*X) + (4*X) + (2*X) + (2*X)\n\n178 = 16 + 8X + 4X\n\n(178-16)/12 = X\n\nX = 13.5
738071
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!)
90547
this must be solved graphically\ny >=-3x+2 , draw this line y=-3x+2 and have the area above it\n\ny<2x+2 , draw y=2x+2 and get the area below the line\n\nThe common region is this 2 areas is the answer