Question

6. (4 pts) Consider the double integral∫R(x2+y)dA=∫10∫y−y(x2+y)dxdy+∫√21∫√2−y2−√2−y2(x2+y)dxdy.(a) Sketch the region of integrationRin Figure 3.(b) By completing the limits and integrand, set up (without evaluating) the integral in polar coordinates.

-1 -2 FIGURE 3. Figure for Problem 6. 6. (4 pts) Consider the double integral V2 /2-y² + = (x2 + y) dx dy + + y) do dy. 2-y2 (a) Sketch the region of integration R in Figure 3. (b) By completing the limits and integrand, set up (without evaluating) the integral in polar coordinates. (x2 + y)dA= / dr do. 7. (5 pts) By completing the limits and integrand, set up (without evaluating) an iterated inte- gral which represents the volume of the ice cream cone bounded by the cone z = 22 + y2 and the hemisphere 2 = 18 - 22 - y2 using (a) Cartesian coordinates. volume = 1 s dz dir dy. (b) polar coordinates. volume = = ] dr de.

Answer 1

I am giving answer of the first question(question number 6) with its two parts . Thank you.

Solution:- √2-12 Vz y Jenay)da = (x²+4) dx dy + (x² + y) dady. R o ay 1 J2-42 For the is the first double integral region given by, (M,y) : -y<n<y, 04 42 | bounded by the lines R, that is =y COA) =-4 (OB) (x-axis) (co म y is the The region graph by region R red giren in coloured Shaded second For the the region R is given double integral by, = 2 2 (2,4): Ja-y2 <2<J2-72, 15*<v? that by the is bounded circle x² + y² - 2

the lines y=1 (CD) y=52 (EF) Region R2 is given in Block coloured shaded Region The the graph region. by the total Hence R is R = Ri R2 in the graph.

Y CA A 1 x=y x=-4 2 E E y=12F y=ly D c4 2 mx too -2 - 1 X! -1 3279²=2 -2 y'

(9) Hence the R is given by R 2 region 3 (2,8) : -J2-g2224 2-y2, yzacy? x² + y2 = 2. That is - 6 in Polar Coordinate x=rcoso. =h rsino 1 7 x² + y2 = 82 =2 [from (8)] => = EN o<r< 52 puo 0284 2T • R = = 20,0): osreſ2 Uz 7 8šo o ( {uz 7 UP = #dodo, x² + y 82 cos²o + 8 sing.

=> x²+y rero cos + sino) s poca - 1 Decembrie r(0 cose +sino) & drdo (rcosho + sino) do do + O O This is the integral in Polar coordinates.