Question
Find the area of the region in the first quadrant enclosed by the coordinate axes and graph of x^3+y^3=1.
Answers
bobpursley
Isnt the height of the function (1-x^3)^1/3 ?
so integrate INT (height)dx from x=0to1
Having a graph(graphing calc) will help understand.
so integrate INT (height)dx from x=0to1
Having a graph(graphing calc) will help understand.
Is it possible to do it by hand? how would I do it by a graphing calculator?
bobpursley
The graphing calculator is to graph the function, so you understand what your integral is supposed to do.
Sure, you can graph it by hand.
Sure, you can graph it by hand.
I cannot find a closed-form solution for the integral in my Tables of Integrals, but the integral can be obtained quite accurately and easily using Simpson's-rule numerical integration. At x values of 0,0.25, 0.5. 0.75 and 1, the values of the function are:
1, 0.995, 0.957, 0.833, and 0
Using the Simpson's Rule formula, I get 0.707 for the integral. Higher accuracy can be obtained by breaking up the 0 to 1 range of integration into larger number of slices. I used four (five data points).
1, 0.995, 0.957, 0.833, and 0
Using the Simpson's Rule formula, I get 0.707 for the integral. Higher accuracy can be obtained by breaking up the 0 to 1 range of integration into larger number of slices. I used four (five data points).
I haven't learned the Simpson's Rule formula yet.
Substitute x = y^1/3 to express it in terms of the Beta function. The integral can then be expressed as:
1/3 Gamma(4/3)Gamma(1/3)/Gamma(5/3) = 0.883319375...
Gamma(x) = (x-1)! and most calculators will evaluate it correctly for fractional values of x.
1/3 Gamma(4/3)Gamma(1/3)/Gamma(5/3) = 0.883319375...
Gamma(x) = (x-1)! and most calculators will evaluate it correctly for fractional values of x.
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