Limits of Integration for y=x^2, Bounded by x=1 and y=1, First Quadrant

[dalarev]

Homework Statement

Given y=x^2 , bounded by the line x=1 and y=1, first quadrant. Fairly simple problem.

Homework Equations

Solving for integration by y first...say, $$\int$$[$$\int$$dy]dx

The Attempt at a Solution

I have solved this problem, integrating by x first and y first. I'm having trouble with a more difficult, yet similar, problem but I feel I should get this fundamental doubt out of the way first.

Integrating by y first, we get the limits of integration from y=x^2 to y=1 . This is rather clear, looking at the graph. The x integration, however, is where I get confused. I would've thought that the limits of x would be from x=0 to x=$$\sqrt{y}$$ because in reality, if you follow the graph, much like we did for the y limits, those are the boundaries x is confined to. The correct x limits, however, are from x=0 to x=1. Any help would be appreciated. Thanks.

 

[benorin]

Starting with y first, we describe the region as being bounded between y=x^2 and y=1, but only the part between x=0 and x=1.

If we instead start with x first, we describe the region as being bounded between $$x=0$$ and $$x=\sqrt{y}$$, but only the part between y=0 and y=1.

 

[dalarev]

Starting with y first, we describe the region as being bounded between y=x^2 and y=1, but only the part between x=0 and x=1.

If we instead start with x first, we describe the region as being bounded between $$x=0$$ and $$x=\sqrt{y}$$, but only the part between y=0 and y=1.

So you mean..starting with y, we think of y as the function part, the dominating part, and x only as the boundaries. When starting with x, x is the boss, the function. y is just the limits, from here to here.

 

[HallsofIvy]

Homework Statement

Given y=x^2 , bounded by the line x=1 and y=1, first quadrant. Fairly simple problem.

On the contrary, it is not a problem at all! I might assume that you forgot to say "find the area" but even so, there is no region bounded by y= x², x= 1, and y= 1. There is a region bounded by y= x² and y= 1. There is another region bounded by y= x², y= 0, and x= 1. Which do you mean?

Homework Equations

Solving for integration by y first...say, $$\int$$[$$\int$$dy]dx

The Attempt at a Solution

I have solved this problem, integrating by x first and y first. I'm having trouble with a more difficult, yet similar, problem but I feel I should get this fundamental doubt out of the way first.

Integrating by y first, we get the limits of integration from y=x^2 to y=1 . This is rather clear, looking at the graph. The x integration, however, is where I get confused. I would've thought that the limits of x would be from x=0 to x=$$\sqrt{y}$$ because in reality, if you follow the graph, much like we did for the y limits, those are the boundaries x is confined to. The correct x limits, however, are from x=0 to x=1. Any help would be appreciated. Thanks.

If the problem is to find the area of the region bounded by y= x² and y= 1, by double integration, using "dy" first, so that the "outer integral" is with respect to x, remember that the limits on the outer integral must be constants. Looking at a graph of y= x² and y= 1, x must range from -1 to 1. Now, for each x, y must range from the graph of y= x² up to the line y= 1. Those are the limits of integration and the integral is
$$\int_{x^2}^1 dydx$$

If, however, the region is bounded by y= x², y= 0, and x= 1, then x ranges from 0 to 1 and, for each x, y ranges from 0 to x². In that case, the integral is
$$\int_0^1\int_{0}^{x^2} dydx$$

Now, what is the region really?

 

[dalarev]

On the contrary, it is not a problem at all! I might assume that you forgot to say "find the area" but even so, there is no region bounded by y= x², x= 1, and y= 1. There is a region bounded by y= x² and y= 1. There is another region bounded by y= x², y= 0, and x= 1. Which do you mean?

You're right, the problem was to find the area bounded by y=x^2 and the line y=1, in the first quadrant only.

If the problem is to find the area of the region bounded by y= x² and y= 1, by double integration, using "dy" first, so that the "outer integral" is with respect to x, remember that the limits on the outer integral must be constants.

I think this just about sums up the problem. The limits on the outer integral must be constants. I suppose the problem here is that I'm not, or wasn't, convinced that that is a fact, I've never seen it stated as a rule or law. But if that's the case, that just makes it easier!

Thanks for the help.