Answer to Question #139340 in Calculus for Promise Omiponle

Question #139340
Using polar coordinates, evaluate the integral ∫∫Rsin(x^2+y^2)dA where R is the region 16≤x^2+y^2≤25.
1
Expert's answer
2020-11-02T18:38:52-0500

Rsin(x2+y2)dA,R={16x2+y225}\iint_R \sin(x^2+y^2)dA, R=\{16\leq x^2+y^2\leq 25\}


x=rcosϕy=rsinϕI=rx = r\cos \phi \\y = r \sin\phi\\ I = r\\

Find the integration limits:

ϕ[0,2π]\phi \in [0,2\pi]

16x2+y22516r2cos2ϕ+r2sin2ϕ2516r2254r516 \leq x^2 + y^2 \leq 25\\ 16 \leq r^2\cos^2\phi + r^2\sin^2\phi \leq 25\\ 16\leq r^2\leq 25\\ 4\leq r \leq 5

r[4,5]r \in[4,5]

So:

Rsin(x2+y2)dA=45dr02πrsin(r2)dϕ==45rsin(r2)drϕ02π=2π45rsin(r2)dr==[r2=u5252rdr=du416]=π1625sin(u)du==πcos(u)1625=π(cos16cos25)==2πsin(412)sin(92)\iint_R sin(x^2 + y^2)dA = \int_4^5dr\int_0^{2\pi} rsin(r^2)d\phi = \\ =\int_4^5rsin(r^2)dr \phi|_0^{2\pi} = 2\pi\int_4^5 rsin(r^2)dr = \\ = \left[ \begin{array}{ccc} r^2 = u & 5 \to 25\\ 2rdr = du& 4 \to 16\\ \end{array} \right] = \pi \int_{16}^{25}sin(u)du =\\ =-\pi cos(u)|_{16}^{25} =\pi(cos16 - cos25) = \\=2\pi sin(\frac{41}{2})sin(\frac{9}{2})


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Comments

Assignment Expert
02.11.20, 00:31

Thank you for correcting us.

Promise Omiponle
31.10.20, 03:34

Hello, you made a big mistake here. The integrated was sin(x^2+y^2) not x^2+y^2

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