1
$\begingroup$

The questions asks:
"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:

My attempt

The Solution:

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Pieter21
    Mar 31 '16 at 9:01
  • $\begingroup$ 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? $\endgroup$
    – abruzzi26
    Mar 31 '16 at 19:30
2
$\begingroup$

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.

enter image description here

$\endgroup$
3
  • $\begingroup$ 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, $\endgroup$
    – abruzzi26
    Apr 1 '16 at 13:34
  • $\begingroup$ I've added a figure. I hope it's useful. $\endgroup$ Apr 1 '16 at 13:50
  • $\begingroup$ Perfect! Thanks you :) $\endgroup$
    – abruzzi26
    Apr 3 '16 at 8:26
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I thought so, the solution graph is incorrect. Thanks for responding $\endgroup$
    – abruzzi26
    Mar 31 '16 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.