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Mathematics

1 answer:

inysia [295]3 years ago

6 0

Oma made a mistake in step B, because she should have divided both sides by b instead of just moving it to the other side. The real next step would look like:
(2A)/b=h

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Evaluate the integral by changing to polar coordinatesye^x dA, where R is in the first quadrant enclosed by the circle x^2+y^2=2

34kurt

$\displaystyle\iint_Rye^x\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta$

which follows from the usual change of coordinates via

$\begin{cases}\mathbf x(r,\theta)=r\cos\theta\\\mathbf y(r,\theta)=r\sin\theta\end{cases}$

and Jacobian determinant

$|\det J|=\left|\begin{vmatrix}\mathbf x_r&\mathbf x_\theta\\\mathbf y_r&\mathbf y_\theta\end{vmatrix}\right|=|r|$

Swap the order, so that the integral is

$\displaystyle\int_{r=0}^{r=5}\int_{\theta=0}^{\theta=\pi/2}r^2\sin\theta e^{r\cos\theta}\,\mathrm d\theta\,\mathrm dr$

and now let $\sigma=r\cos\theta$ , so that $\mathrm d\sigma=-r\sin\theta$ . Now, you have

$\displaystyle\int_{r=0}^{r=5}\int_{\sigma=0}^{\sigma=r}re^\sigma\,\mathrm d\sigma\,\mathrm dr=\int_{r=0}^{r=5}r(e^r-1)\,\mathrm dr=4e^5-\dfrac{23}2$

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3 years ago

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