I need to change the order of integration and evaluate the following:
$\int_{0}^{1} \int_{\sqrt{y}}^{1} e^{x^2}dxdy$
$x=\sqrt{y}, x=1$
Would this work: $\int_{0}^{1} \int_{0}^{x^2} e^{x^3}dydx$ ?
How change the order and integrate this function $e^{x^{2}}$
integrationmultivariable-calculus
Related Solutions
In your first integral, your region is bound by
$x= \sqrt {\frac y2}\\ x = 9\\ y = 0\\ y = 324$
Graphing of these curves
We get two regions.
If this is indeed what you are trying to integrate.
flipping the order of integration
$\int_0^9 \int_0^{2x^2} f(x,y) \ dy \ dx + \int_9^{9\sqrt{2}}\int_{2x^2}^{324} f(x,y) \ dy \ dx$
Or, as Daniel Fisher points out, your curve is supposed to be $x= \frac {\sqrt y}{2}.$
$\int_0^9 \int_0^{4x^2} f(x,y) \ dy \ dx $
You can reverse the order but the limits of integration change:
$$\int_{y=0}^{\sqrt{2}}\left(\int_{x=y^2}^{2}y\ dx\right)dy=\int_{x=0}^{2}\left(\int_{y=0}^{\sqrt{x}}y\ dy\right)dx.$$
Best Answer
The original domain is $0 \le y \le 1, \sqrt{y} \le x \le 1$
It is equivalent to $0 \le x \le 1, 0 \le y \le x^2$.
Hence the part about changing the variable is correct, however, note that the function that you are integrating should remain the same.
\begin{align} \int_0^1 x^2 \exp(x^2) \, dx &= \int_0^1 \frac{x}2 (2x) \exp(x^2) \, dx \\ &= \frac{x}2 \exp(x^2) |_0^1 - \int_0^1 \frac12 \exp(x^2) \, dx \\ &= \frac{e}2 - \frac12 \int_0^1 \exp(x^2) \, dx \\ &= \frac{e}{2} - \frac{\sqrt{\pi}}{4}erfi(1) \end{align}