2006-09-03 00:21:15 · 4 answers · asked by jhen_hidaka 1 in Science & Mathematics ➔ Mathematics
Assuming this is the area bounded by the curve y= 9 - x^2 and the x-axis, the area,
A= integral from -3 to 3, [9 - x^2] dx
= [9x - (x^3)/3] from -3 to 3
= (27 - 9) - (-27 + 9)
= 54 -18
= 36 square units
This is a definite integral and the boundary has been determined by considering the points at which the curve cuts the x-axis,i.e
y=0 => 9 - x^2 = 0 => x = 3 or -3
2006-09-03 04:05:54 · answer #1 · answered by yasiru89 6 · 0⤊ 0⤋
Do you mean :"the area bounded by y= 9-x^2 and x-axis"?
y=9-x^2 is the formula of a upside down parabola which cross the Y axis at 9. The area between y axis and the parabola is infinite!
2006-09-03 08:18:09 · answer #2 · answered by Farshad 2 · 0⤊ 1⤋
integral from -3 to 3 of (-x^2+9 ) dx
or
2 times of integral from 0 to 3 of (9-x^2)dx
it will be 2(9x-x^3/3) from 0 to 3
the value is 27-9=18 is the answer
2006-09-03 07:40:56 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
Interchange x & y [NOTE THAT QUESTION IS NOT BEING CHANGED] So it becomes y^2=(9-x) & x axis
We can integrate now
But we notice that it is symmetric hece for every +ve area there is same corresponding -ve area hence zero
If you want to find modulus it is infinite
2006-09-03 09:48:10 · answer #4 · answered by Love to help 2 · 0⤊ 0⤋