I have a problem I have been working on, with the solution but the thing is I don't really understand how it is done.

The question, is to compute, $$\int_0^1 \int_{9x^2}^9 x^3\sin(8y^3) \,dy\,dx $$

Now, I did notice that we are going to have to reverse the order of integration so first I took note of, as of now I have $$0 \le x \le 1$$ and $$9x^2 \le y \le 9$$ and I tried to consider the graph. This is where I am getting confused, I don't know if I am supposed to consider the area basically above the line $$0\le x\le\sqrt{\frac{y}{9}}$$ and put $0 \le y \le 9$ and compute. I know that is what I should do, but I am having a lot of trouble seeing this from the graph. My apologizes as I am not aware of how to put graphs on the site.

I mean I am having trouble visualizing what it is meant to say $x$ is less than that value of $y$, when are we not considering the region bounded above?

I appreciate all answers and comments, ideally though I would like an answer that includes graphics if possible!

Could anyone shed some light on this? Ps, this is not homework and I already have the final solution if anyone wants to check, it is $$=\frac{1-\cos(5832)}{7776}$$

Thank you all

  • $\begingroup$ See math.stackexchange.com/questions/1073275/… - it explains this all. See... you've not actually got two integrals, you have $\int_Sf(p)dA$ - a surface and you're integrating (summing over (it's a comment not quite summing but lets not be pedantic!)) over little chunks of area. That answer should help you $\endgroup$ – Alec Teal Apr 9 '15 at 22:46
  • $\begingroup$ To put a gap between this and the above comment, your area is x from 0 to 1, and for each x you go from $9x^2$ to $9$, so your surface is the region of all the points between the lines $y=0$, $x=0$, $x=1$ and the curve $y=9x^2$ you can parameterise that however you like! $\endgroup$ – Alec Teal Apr 9 '15 at 22:49

Since you would like an answer that includes a graph, I've prepared the following simple one.

enter image description here

Your double integral

\begin{equation*} I=\int_{0}^{1}\int_{9x^{2}}^{9}x^{3}\sin \left( 8y^{3}\right) \,dy\,dx \end{equation*}

is to be evaluated in the region $R$ of the 1.st quadrant bounded from below by the graph of the function $y=9x^2$, from above by the horizontal line $y=9$, and from left by the vertical line $x=0$, because the inequalities you have found, $0 \le x \le 1$ and $9x^2 \le y \le 9$, define exactly $R$, i.e.

\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq 1,\; 9x^{2}\leq y\leq 9\right\}. \end{equation*}

How to define the very same region with a different pair of inequalities? Since $y=9x^2$ is equivalent, for $x\ge 0$, to $x=\sqrt{y/9}=\sqrt{y}/3$, it is clear from the picture that you can also define it by the inequalities $0\leq x\leq \sqrt{y}/3$ and 0$\leq y\leq 9$, as you have done. In symbols,

\begin{equation*} R=\left\{ (x,y)\in \mathbb{R}^{2}:0\leq x\leq \sqrt{y}/3,\; 0\leq y\leq 9\right\}, \end{equation*}

which results in the integral

\begin{equation*} I=\int_{0}^{9}\int_{0}^{ \sqrt{y}/3}x^{3}\sin \left( 8y^{3}\right) \,dx\,dy. \end{equation*}


In 2D I usually prefer to do this by drawing a picture. Let me talk through how I would actually draw this picture.

The region $0 \leq x \leq 1,9x^2 \leq y \leq 9$ is bounded by the lines $x=0,y=9$ and the curve $y=9x^2$. You can think of the limits that you have as saying that for each $x$ between $0$ and $1$, you integrate from $9x^2$ to $9$ in $y$. So this corresponds to little vertical segments between the parabola and the horizontal line $y=9$.

Now you want to go the other way: for each value of $y$ between $0$ and $9$, what are the possible values of $x$ in the region? These now correspond to little horizontal segments, which must be between the vertical line $x=0$ and the parabola $y=9x^2$. Solving for $x$ you get that the parabola is given by $x=\sqrt{y/9}$, so for each $y$, $x$ ranges from $0$ to $\sqrt{y/9}$. So your new integral looks like $\int_0^9 \int_0^{\sqrt{y/9}} f(x,y) dx dy.$


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