# Area of region bounded by two parabolas using double integrals

"Let R be the region in the first quadrant bounded by the graphs of the parabolas $y=2x^2$, $y=9-x^2$ and the line x=0. Express the area of region R:
(i) Integrating first with respect to y, and then with respect to x
(ii) Integrating first with respect to x, and then with respect to y "

I have tried sketching the two curves, and expressing the double integral. However, my sketch and the double integral itself is quite different from the solution. Please have a look at my attempt and the solution below:

The Solution:

Could someone please help me understand this solution? I feel like I am missing a very important concept about double integrals. Any help would be highly appreciated.

• The area that you drew is not correct. It should extend upward until the $9-x^2$ parabola. And that is only one of two solutions. Mar 31 '16 at 9:01
• Thanks for the reply. But after extending the graph, I would still get a similar shape to the one I currently have. The solution's graph is different. Could it be incorrect? Mar 31 '16 at 19:30

If I well understand your question maybe that this intuitive interpretation can help.

When we calculate an area by a double integral, we subdivide this area in a sum of little ''area elements''. If we chose these elements as $dy dx$ or $dx dy$, this means that the elements are little rectangles of sides $dy$ and $dx$.

If we chose the order $dydx$ this means that we want to count these elements starting from ''vertical'' strips in which (in your case) the side $dy$ start from $2x^2$ and goes to $9-x^2$, so these values are the limits of the sum, and becomes the limits of the integration in $dy$, than we sum all these strips summing for the oter side $dx$ has limits $o$ and $\sqrt{3}$, and this gives the double integral: $$\int_0^{\sqrt{3}}\int_{2x^2}^{9-x^2}dydx$$

If we change the order to $dx dy$ we count the area elements starting from horizontal stripes, and in this case the side $dx$ start from $x=0$ and goes to $\sqrt{y/2 }$ if $y\le 6$, and to $\sqrt{9-y }$ if $6<y\le 9$. So you have the other double integral that is divided in two parts.

• Thank you very much for this explanation; it makes perfect sense to me. But how do I go about sketching the projection? I have a habit of always sketching the region before setting up the double integrals, Apr 1 '16 at 13:34
• I've added a figure. I hope it's useful. Apr 1 '16 at 13:50
• Perfect! Thanks you :) Apr 3 '16 at 8:26

The first graph you draw is good, but you paint incorrectly. You should have painted from left: $x=0$, from up: $y = 9-x^2$, and from down: $y = 2x^2$.

The graph in the solution is not correct, i think, but the double integrals are true.

• I thought so, the solution graph is incorrect. Thanks for responding Mar 31 '16 at 19:31