By using the double integrals,find the area of regions enclosed by curve $y=5-x^2$ and line $y=x+3$
Here is the problem, I've done sketching the graph but I dont know how to solve it . Can someone help me. Thanks
Double integrals for area region
areacalculusintegration
Related Solutions
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.
You can start from $dA=dxdy$. To compute the area, you need to find $$A=\int_Ddx\,dy,$$ where $D$ is the region you are interested in. Now let's try to describe this region $D$. You can cut this region into vertical slices. When $x$ is fixed, $y$ will grow from $2-x$ to $\sqrt{4-x^2}$. So the "height" of the region at this point $x$ is $\int_{2-x}^{\sqrt{4-x^2}}\,dx$. Then you can integrate over $x$. So you will have \begin{align*} A&=\int_D\,dx\,dy\\ &=\int_0^2\left(\int_{2-x}^{\sqrt{4-x^2}}\,dy\right)\,dx\\ &=\int_0^2\left[\sqrt{4-x^2}-(2-x)\right]\,dx. \end{align*} This can be solved by 1d-calculus.
Best Answer
It's easy to see that the two intersection points of $y=5-x^2$ and $y=x+3$ are $(-2,1)$ and $(1,4)$.
So the region bounded by the two curves can be defined as $$E:=\{(x,y)\in\mathbb R^2:3+x\le y\le 5-x^2, x\in[-2,1]\}\implies$$ $$\text{Area}(E)=\int_{-2}^1\int_{3+x}^{5-x^2}dydx=\int_{-2}^1(2-x^2-x)dx=2x-\dfrac{x^3}{3}-\dfrac{x^2}{2}\bigg\vert_{-2}^1$$