areacalculusgeometry
Namely we have given two graphs: $y=\sqrt{2 – x^2} \text {and } x + (\sqrt{2}-1)y = \sqrt{2}$
The question asks us to find the area bounded by those two graphs. I tried sketching those two graphs in a plane, however still I'm not sure which area is the question asking to find?
The correct answer says $\frac{\pi }{4} – \frac{1}{\sqrt{2}}$. Here is a picture of the two graphs. I would be really thankful if someone shows me how to approach to find the correct area.
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.
Yes, correct as you had found. $$\int_\limits{0}^9\sqrt{y} dy = 18$$ $$\dfrac{18}{2}=\int_\limits{k}^9\sqrt{y}dy \rightarrow k=.. $$
Best Answer
I would not think about graphs, but rather think about circles and lines in the plane instead.
The sector of the circle defined by the intersections with the line (the shaded region in the drawing above) are an 8th of the circle. So the whole sector has area $\frac18\cdot \pi\sqrt 2^2 = \frac\pi4$.
Now, this sector includes the region we are after, but also includes a triangle $\triangle ABO$ that we don't want. So we subtract the area of that triangle. It has base $\sqrt2$ and height $1$, so its area is $\frac12\cdot \sqrt2 = \frac1{\sqrt2}$. This gives you the answer you're after.