mirandasatterley said:
Homework Statement
Find the volume of the region under the graph of f(x,y) = x+y and above the region y2≤x, 0≤x≤9
The Attempt at a Solution
From these equations, x will be integrated from 0-9, but I'm not sure about y.
My thinking is that y will be intgrated from 0-3 because y2≤x and the smallest value of x is 0, and the square root of 0 is 0, so that is the smallest y, and the largest x value possible is 9 and the positive quare root of 9 is 3, so this is the largest value of y,
So I would integrate:
∫0-9∫0-3 (x+y)dydx.
Is this correct or am I missing something, where the limits of integration also involve using f(x,y)?
No, that is not correct. The region a\le x\le b, c\le y\le d, with a, b, c, d numbers is always a
rectangle and the figure here is not a rectangle. But the bounds do NOT involve f(x,y)- that is a "z" value and goes inside the integral as you have it.
Always draw a picture for problems like this. y^2= x is a parabola, of course, "on its side". The line x= 9 is a vertical line crossing the parabola at (9,3) and at (9, -3). Yes, you can integrate with x going from 0 to 9. On your picture, mark an arbitrary "x" by marking a point on the x-axis between 0 and 9. Now draw a vertical line from one boundary to the other. The y bounds,
for that x, are y values of those endpoints: (x, -\sqrt{x}), and (x, -\sqrt{x}). Your integral is
\int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}} f(x)dy dx= \int{x=0}^9\int_{y=-\sqrt{x}}^{\sqrt{x}}(x+ y) dy dx
Of course, like any double integral, you can reverse the order of integration. If you look at your picture you will see that y ranges, overall, from -3 to 3. Draw a horizontal line across the parabola, representing an arbitrary value of y in that range. It will have left endpoint on the parabola: x= y^2, and right endpoint on the vertical line x= 9. Those will now be the limits of integration for this order:
\int_{y= -3}^3\int_{x= y^2}^9 f(x,y)dx dy= \int_{y= -3}^3\int_{x= y^2}^9(x+ y)dx dy
Try it both ways. You should get the same answer.
