Answer to Question #140335 in Calculus for Promise Omiponle

Question #140335
Evaluate ∫∫Rcos(x^2+y^2)dA, where R is the region above the x-axis within the circle x^2+y^2= 25.
1
Expert's answer
2020-11-05T10:33:18-0500

We convert to polar co-ordinatesdA=rdrdθx=rcosθy=rsinθx2+y2=r2SinceRis the region above thexaxis,we only consider the first andsecond quadrants, and thus evaluatethe double integral.θis from0toπris from0to5Rcos(x2+y2)dA=050πcos(x2+y2)rdθdr=π05cos(r2)rdr=πsin(r2)205=πsin(25)2The area of the region isπsin(25)2\displaystyle \textsf{We convert to polar co-ordinates}\\ \mathrm{d}A = r\mathrm{d}r\mathrm{d}\theta\\ x = r\cos\theta\\ y = r\sin\theta\\ x^2 + y^2 = r^2\\ \textsf{Since}\, R\, \textsf{is the region above the}\, x-\textsf{axis},\\\textsf{we only consider the first and}\\\textsf{second quadrants, and thus evaluate}\\\textsf{the double integral.}\\ \theta\, \textsf{is from}\, 0 \, \textsf{to}\, \pi\\ r\, \textsf{is from}\, 0 \, \textsf{to}\, 5\\ \begin{aligned} \therefore \iint_R \cos(x^2 + y^2) \, \mathrm{d}A &= \int_0^5\int_0^\pi \cos(x^2 + y^2) \, r\mathrm{d}\theta \mathrm{d}r\\ &= \pi\int_0^5 \cos(r^2) \, r \mathrm{d}r\\ &= \frac{\pi\sin(r^2)}{2} \biggr\vert_0^5\\ &= \frac{\pi\sin(25)}{2} \end{aligned}\\ \textsf{The area of the region is}\,\frac{\pi\sin(25)}{2}


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