definite integralsintegrationmultivariable-calculus
Sketch the region of integration and evaluate the integral:
$$\int_1^2 \int_y^{y^2} dx \, dy$$
I understand how to take the integral, but the region of integration seems like it has no bounds. Like between y=1 and y=2, the graphs of $y = \sqrt{x}$ and $y = x$ diverge, and don't form a closed region.
You need to be more clear about your double integral. Say you have $$ \int_c^d \left(\int_a^b f(x,y)g(x,y)dx\right) dy$$ And you need to know the antiderivative of $g(x,y)$ with respect to $x$. So the information $\int_X g(x,y)dx=w(y)$ is not enough. Because this is not an antiderivative of $g$ with respect to the $x$ direction. Instead, you need to have $$ \int_a^x g(s,y)dy=h(x,y)$$ That is, $$ \int_a^b g(x,y)dy=h(b,y)-h(a,y)=h(b,y)=w(y)$$ Under this assumption, you can mimic the way how we do integration by parts in 1 dimensional. $$ \int_c^d\left(\int_a^b f(x,y)g(x,y)dx\right)dy=\int_c^d\left(f(x,y)h(x,y)\Big|_a^b-\int_a^b h(x,y)\frac{\partial}{\partial x}f(x,y)dx\right)dy$$ $$ =\int_c^d f(b,y)w(y)dy-\int_c^d\left(\int_a^b h(x,y)\frac{\partial}{\partial x}f(x,y)dx\right)dy$$
If you know the antiderivative of $h(x,y)$ with respect to $y$, i.e. $$ \int_c^y h(x,t)dt=\int_c^y\int_a^x g(s,t)ds\, dt=l(x,y)$$ you can keep going: $$ \int_c^d\left(\int_a^b f(x,y)g(x,y)dx\right)dy=\int_c^d f(b,y)w(y)dy-\int_c^d\left(\int_a^b h(x,y)\frac{\partial}{\partial x}f(x,y)dx\right)dy=\int_c^d f(b,y)w(y)dy-\int_a^b\left(l(x,d)\frac{\partial}{\partial x}f(x,d)-\int_c^d l(x,y)\frac{\partial^2}{\partial x\partial y}f(x,y)dy\right)dx=...$$
It all depends on what information you have for $f(x,y)$ and $g(x,y)$.
If you want to integrate first in order to $y$ and then in order to $x$, then start by noting that $0\leqslant x\leqslant\sqrt2$ (see the picture below). If $0\leqslant x\leqslant1$, then $y$ takes values from $-\sqrt{2-x^2}$ to $-\sqrt{1-x^2}$ and from $\sqrt{1-x^2}$ to $\sqrt{2-x^2}$, whereas if $1\leqslant x\leqslant\sqrt2$, then $y$ takes values from $-\sqrt{2-x^2}$ to $\sqrt{2-x^2}$. So, you have$$\int_0^1\left(\int_{-\sqrt{2-x^2}}^{-\sqrt{1-x^2}}x^2\,\mathrm dy+\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dy\right)\,\mathrm dx+\int_1^{\sqrt2}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dy\,\mathrm dx.$$
And if you integrate first in order to $x$ and then in order to $y$, by a similar reason you should get$$\int_{-\sqrt2}^{-1}\int_0^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy+\int_{-1}^1\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy+\int_1^{\sqrt2}\int_0^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy.$$
Best Answer
The region of integration is in fact bounded. First, we integrate with respect to $x$ over the interval of integration $[y,y^2]$. It's true that $y$ and $y^2$ diverge as $y\rightarrow\infty$. However, the bounds on the second integration w.r.t. $y$ are only from $y=1$ to $y=2$. Together, the region of integration in the $xy$-plane can be represented mathematically by the pair of inequalities:
$$y \leq x \leq y^2,\\ 1 \leq y \leq 2.$$
Note that since $1 \leq y \leq 2$, it follows that $1 \leq y^2 \leq 4$. Going back to the first inequality with this bounds in mind,
$$1 \leq y \leq x \leq y^2 \leq 4$$.
Thus, the region of integration must be a subset of the rectangle $[1,4]\times[1,2]$.