Question

Use polar coordinates to evaluate the integral

SUR (60 - 3y)dA

where R is the region in in the first quadrant enclosed by the circumference x2+y2=4 and the lines x=0 and y=x

Answer 1

by x=0 We have the integral dA where R is the region in the 1st : RE quadrant bounded by 22ty? 4 and. (y-axis) and y=x Let us consider the substitution x= p loso Yarsino Now,. On the circle apy? =4, we have 8=2 and on the we have on line you OaT/4

Now, at the origin we must have p=0 and on the y-axis 0= 1/2 because on the y-axis so Cosco 20 =), 03149 (***). 10, esinya) x=0. i e, (relost, psin II 9 Hore OLr 22 and y=k N=0 7430 120 €/ and dA=p.dr de So, we have 2 from on the given integral T/2 ro Cos 0. – Prosine) (6 a da die T/ dro - co I (6rtime + 370630) * {bomºto) - (San ma?)}ds ( 6 Iru D 2 1. S + 322 o

si (66² - 982 dr a 23 6- X 9 ra 3 a x (6-9 v2 16 24 r2 16. 1282 square unit Hence, the required area is. (16-12-2) sq- wnits 3rd option is connect.

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