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Evaluate the double integral $\displaystyle \iint_D 2x\cos\!\left(y\right)+5 \ dA $over the region D which is bounded by $y = 0, y = 2x^{2} $and $x = 1.$

$\displaystyle \iint_D 2x\cos\!\left(y\right)+5 \ dA = \int_a^b\int_c^d 2x\cos\!\left(y\right)+5 \ dy \ dx$

a = 0, b = 1

c = 0, y = 2(?)

why is y not equal $2$?

$y = x^2$ and $x = 1$ intersect when $y = 2$

Wouldn't the double integral be:

$\int_0^1\int_0^2 2x\cos\!\left(y\right)+5 \ dy \ dx$

p.s. here's the screenshot

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    $\begingroup$ If you let $x:0 \to 1$ and $y:0 \to 2$, you're integrating over a rectangle. Make a sketch of the region $D$ to see that this is not what you want. $\endgroup$
    – StackTD
    Mar 11 '16 at 16:15
  • $\begingroup$ @StackTD well I did, link, drawn on a 2d plane, and then you can imagine that whatever surface is the red region, it just extends towards you and beyond (z-axis). No? $\endgroup$
    – Jack
    Mar 11 '16 at 16:48
  • $\begingroup$ Yes, but you don't have to visiualize the function you're integrating to set up the integral boundaries: all you need is a good view of the (red) region $D$. I'll refer to your sketch in an answer. $\endgroup$
    – StackTD
    Mar 11 '16 at 16:49
  • $\begingroup$ I elaborated in an answer; hope this helps. $\endgroup$
    – StackTD
    Mar 11 '16 at 16:59
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It is a good idea to let $x$ run from $0$ to $1$. Referring to your sketch of the region $D$, you then need to find the limits for $y$ so that you run exactly through the red area: no more and no less.

Position yourself at an arbitrary $x$, somewhere between $0$ and $1$, below the $x$-axis. Let $y$ increase: you encounter the (red) region $D$ at $y=0$, this is where you start integrating. You leave the region when you cross the blue curve, corresponding to $y=2x^2$. This is where you stop integrating. So you end up letting $y$ run from $0$ to $2x^2$.

Continuing to $y=2$ (you only want this at $x=1$, but not for all the $x$ values $0 \le x < 1$!) would give you a complete rectangle and this is not what you want.

It should be clear from your sketch that the upper boundary of the region is not a constant (like $2$), but a function of $x$.

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  • $\begingroup$ I think I'm getting it... but how come, so far when setting boundaries for y... I just figured out what the rightmost/leftmost x point corresponds in y -axis, and that's it.... $\endgroup$
    – Jack
    Mar 15 '16 at 3:43
  • $\begingroup$ I don't fully understand your question, I think... Perhaps you can clarify? In any case: for a region like this, you can choose wether you take fixed limits for $x$ or for $y$. But then you will have variable limits for the other one (as a function of the fixed, outer one) because with fixed limits for both, you'll always integrate over a rectangle (sketch this!). The boundary functions determine what these variable limits will be and when they're given in the form $y=f(x)$, it's usually the easiest to take fixed limits for $x$ and the variable ones for $y$. $\endgroup$
    – StackTD
    Mar 15 '16 at 8:41

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