lalah said:
Homework Statement
I'm trying to reverse the order of integration of
\int_0^{81}\int_{y/9}^{\sqrt{y}} dx dy
first integral is from 0 to 81
second integral is from y/9 to \sqrt{y}
The problem is, I identified the inequalities for the regions of integration, but I'm having problems understanding how the graph is formed.
Homework Equations
The Attempt at a Solution
I understand the inequalities are y/9 < x < \sqrt{y}
and 0 < y <81.
But I don't understand the graph.
x-axis is 0 to 9
y-axis is 0 to 81
y = x^2
y = 9x
And since according to the homework problem I'm doing, I can't progress the problem until I understand why the graph is so.
Also, can someone help me in formating the integrals in latex?
Draw a picture. In particular, graph x= 1/y and x= \sqrt{y} which are the same as the hyperbola y= 1/x and the parabola y= x
2. Now one problem you have here is that those two graph cross at (1,1). The area over which you are integrating is the area from the curve y= x
2 on the left to the curve y= 1/x on the right up until y= 1, then from the curve y= 1/x on the left up to the curve y= x
2 on the right.
Reversing the order of integration, you will need to divide this into
four integrals. To cover the lower area, below y= 1, because of the change at (1,1), you will need to integrate from 0 up to y= x
2 with x from 0 to 1, then from 0 up to y= 1/x, with x from 0 to 1/81. To cover the upper area, above y= 1, you need to integrate from y= 1/x up to y= 81, with x from 1/81 to 1, then from y= x
2 up to y= 81, with x from 1 to up to 9:
\int_{x=0}^1\int_{y= 0}^{x^2} dy dx+ \int_{x=1}^\infty \int{y= 0}^{1/x} dy dx+ \int_{x= 1/81}^1 \int_{y=1/x}^81 dy dx+ \int_{x=1}^9\int_{y= x^2}^{81} dy dx
