double integral

**Dr Zoidburg** · May 4th 2010, 06:17 PM

I = ∫∫R (x + y)dA where R is the region bounded by the curve y = x² and the line y = 1.
(a) Calculate I by integrating first in y and then in x.
(b) Calculate I by integrating first in x and then in y.

This is what I got for (a): Limits for y are x² =< y =< 1, making this a parabola from origin to y=1. solving x²=1 (upper limit of y) gives the limits for x -1=<x=<1.

$\int_{-1}^{1}\int_{x^2}^{1} (x+y)$

$\int_{-1}^{1} (xy+ \frac{y^2}{2})\bigg|_{x^2}^{1}$

$\frac{1}{2} \int_{-1}^{1} (-x^4 - 2x^3 + 2x +1)$

$\frac{1}{2} (\frac{-x^5}{5} - \frac{x^4}{2} + x^2 +x)\bigg|_{-1}^{1}$

$=\frac{4}{5}$

which I hope is correct!

I'm stuck on (b)
What should I make the limits of x & y to?
are the limits for x: $-1 \leq x \leq \sqrt{y}$ ? This is from solving x² = y, making $x=\sqrt{y}$
But what do I make the limits for y? If I use $0 \leq y \leq 1$ I get a different answer from (a). I get 13/20.

where am I going wrong?

**drumist** · May 4th 2010, 06:50 PM

$x^2=y ~\implies~ |x| = \sqrt{y} ~\implies~ x = \pm \sqrt{y}$

So, this means that $x=\sqrt{y}$ where x is positive $x=-\sqrt{y}$ where x is negative.

Therefore, the endpoints will be

$\int_0^1 \int_{-\sqrt{y}}^{\sqrt{y}} (x+y) \, dx \, dy$

**Dr Zoidburg** · May 4th 2010, 07:17 PM

D'oh! It's so obvious. Thanks heaps for that. It was bugging me no end I've got this assignment due in a couple of days.

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