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The question is:

integrate $f(x,y) = 1$ on the region bounded by:

$$0 \le y \le -x^2 + 1$$ $$-1 \le x \le 0.5$$

I've turned this into the following integral:

$$\int_0^1 \int^{0.5}_{\sqrt{1-y}} 1 dx dy$$

When I evaluate this, I get $\frac{1}{6}$. However, this is equivalent to the 1D integral:

$$\int_{-1}^{0.5} (-x^2 + 1)dx$$

which correctly evaluates to 1.125.

Clearly, my bounds are a problem, but I can't figure out where I went wrong.

Michael Stachowsky
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1 Answers1

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Try $$\int_0^{0.75} \int^{0.5}_{-\sqrt{1-y}} 1 dx dy $$ + $$\int_{0.75} ^1\int^{\sqrt{1-y}}_{-\sqrt{1-y}} 1 dx dy $$

Indeed, x=0.5 intersects the parabola in y=0.75

MysteryGuy
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    Shouldn't the upper bound on the internal integral be 0.5? (I suspect yes, since with that bound we get the correct answer :-) ). I see what I did wrong, though. I was integrating on the wrong part of the parabola. Thanks! – Michael Stachowsky Apr 20 '17 at 17:11
  • @MichaelStachowsky Yes, you're right. I've edited my answer so now it's correct ;) – MysteryGuy Apr 20 '17 at 17:20
  • Actually, I don't think it is. If we use those bounds, we end up with the value 1.333. On the other hand, if we do the 1D integration, we get 1.125. I believe the challenge here is that we need to consider several sub-domains, since the bounds are not monotonic and thus don't have inverses – Michael Stachowsky Apr 20 '17 at 17:24
  • @MichaelStachowsky this last should be great :-) – MysteryGuy Apr 20 '17 at 17:56
  • Yes, that did it, thanks – Michael Stachowsky Apr 20 '17 at 17:58