Question

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=25

Answer 1

`\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-\frac{23}2`