2007-11-26 06:18:49 · 5 answers · asked by Andy 2 in Science & Mathematics ➔ Mathematics
write the double integral that calculates the area bounded by these graphs:
y=x
y=9/x
x=9
x=0
y=0
2007-11-26 06:24:20 · update #1
area = ∫{0, 9}[∫{x/9, x} dy] dx
2007-11-26 06:29:21 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
There's a little problem with your 5 graphs. The y = x, x=9, x=0 and y= 0 makes an isosceles triangle in quadrant 1, then the y= 9/x cuts thru that triangle dividing it into 2 parts.
Seem like it might be a trick question where the simple answer is the area of the isosceles triangle 1/2 (9)(9) = 81/2,
and the graph of y=9/x is irrelevant.
I THINK that sahsjing is getting the area uppermost in the right isosceles triangle and I presume that the quantities in the braces are limits {lower, upper} for the integral. If so, then the outer integral's limits would be [3,9].
The upper region has an area of 32 sq units. The lower region touching the x-axis and origin has an area 8.5 sq units.
2007-11-26 14:54:22 · answer #2 · answered by answerING 6 · 0⤊ 0⤋
What's ur question?
2007-11-26 14:22:42 · answer #3 · answered by tinhnghichtlmt 3 · 0⤊ 0⤋
iyiogrenci@yahoo.com
2007-11-26 14:23:13 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋
I can!
2007-11-26 14:21:31 · answer #5 · answered by Anonymous · 0⤊ 0⤋