Establishing double integral limits

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PedroB
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Homework Statement



What would be the limits for each of the integrals (one with respect to x, one with respect to y) of an area bounded by y=0, y=x and x^2+y^2=1?


Homework Equations



None that I can fathom

The Attempt at a Solution



I've rearranged the latter most equation to get x=√(1-y^2) and tried subbing in values for y=0 and y=x but that does not seem to work correctly.

If it helps this is part of a problem that involves evaluating an integral by first changing the variables from x and y to u and v; while I am given what the new limits are in terms of u and v I cannot seem to be able to work backwards to obtain them in terms of x and y.
 
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PedroB said:

Homework Statement



What would be the limits for each of the integrals (one with respect to x, one with respect to y) of an area bounded by y=0, y=x and x^2+y^2=1?

Homework Equations



None that I can fathom

The Attempt at a Solution



I've rearranged the latter most equation to get x=√(1-y^2) and tried subbing in values for y=0 and y=x but that does not seem to work correctly.

If it helps this is part of a problem that involves evaluating an integral by first changing the variables from x and y to u and v; while I am given what the new limits are in terms of u and v I cannot seem to be able to work backwards to obtain them in terms of x and y.

Another words, you want to compute the area in carteasian coordinates and to avoid doing two double integrals, integrate with respect to x first, then y so need to compute:

[tex]\int_{y=y_0}^{y=y_1} \int_{x=g_1(y)}^{x=g_2(y)} 1 dxdy[/tex]

You drew a picture yet? I did and x is going from that diagonal line to the circle. Need to get both of those in terms of x(y). The diagonal line is easy. It's [itex]g_1(y)=x(y)=y[/itex]. You can express the ending limit on x by expressing x as a function of y for the equation of the circle. That gives you the inner integral. Now what is the limits on y? Well, y is going from zero to the point where the diagonal line hits the circle. You can finish it.

Edit: Oh, I see what haruspex is saying. I'm assuming it's below the diagonal line in the first quadrant. If not sorry.