Evaluate the surface integral// (-si + 3zk) - as.Where S consists of the paraboloid y = 2 + z2,0< y< 16 and the disk + z2 < 16, y = 16, and has outward orientation.Consider splitting the surface into its parts S = S1 U S2 where Si is the paraboloid and S2 is the disk. Then we haveF. dS =F. dS, +F. dS2Si can be parametrized by ri (s, t) = (s cos (t), s^2Σssin(t)S2 can be paraetrized by r2 (s, t) =(s cos (t), 16Σssin(t)(-si + 3zk) · ds =(s^2+3ssin(t)"sqrt(4s^4+s*2)E ds dt +|-((s^2+3ssin(t)"sqrt(4s^4+s^2))/2Σds dt(Evaluate S, for the first blank and S2 for the second blank)wheret1 =s1 =Σto = 2piΣs2 =4ΣEvaluate(-si + 3zk) · dS =4"4pi+ -4^4pi/2Σ(Evaluate S1 for the first blank and S2 for the second blank)M MM M

To find- Evaluate the surface integral ∬S-yj + 3zk·dS where S consists of the paraboloid y = x2 + z2, 0 ≤ y ≤ 16 and the disk x2 + z2 ≤ 16, y = 16, and has outward orientation.Explanation- Consider the Surface S = S1 ∪ S2whereS1 is the paraboloidS2 is the disk Now, S1 : y = x2 + z2, 0 ≤ y ≤ 16Letx = scostz = ssintSo,y = scost2 + ssint2 = s2cos2t + s2sin2t = s2cos2t + sin2t = s2⇒y = s2So,S1 can be parametrized by r1s, t = s cost, s2, s sint Now, S2 : x2 + z2 ≤ 16, y = 16Letx = scostz = ssintSo,y = 16⇒s2 = 16⇒s = 4So,S2 can be parametrized by r2s, t = s cost, 16, s sintorS2 can be parametrized by r2s, t = 4cost, 16, 4sintNow, Consider S1 : f = y - x2 - z2 So, Gradient of f is: fx, fy, fz = -2x, 1, -2z And Unit normal vector = --2x, 1, -2z-2x, 1, -2z = 2x, -1, 2z-2x, 1, -2z So, The integral becomes ∬S-yj + 3zk·dS = ∬S1-yj + 3zk·2x, -1, 2z-2x, 1, -2z·-2x, 1, -2zdS1 = ∬S1-yj + 3zk·2x, -1, 2z dS1 = ∬S1y + 6z2 dS1 = ∬S1x2 + z2 + 6z2 dS1 = ∬S1x2 + 7z2 dS1⇒∬S-yj + 3zk·dS = ∬S1x2 + 7z2 dS1 .........1 Now, For S1 The limits of t are: 0 ≤ t ≤ 2π The limits of s are: 0 ≤ s ≤ 4 So, Equation (1) becomes ∬S1x2 + 7z2 dS1 = ∫02π∫04scost2 + 7s sint2s dsdt = ∫02π∫04s2cos2t + 7s2sin2ts dsdt⇒∬S1x2 + 7z2 dS1 = ∫02π∫04s2cos2t + 7s2sin2ts dsdtNow, Consider S2 : f = y = 16 So, Gradient of f is: 02 + 12 + 02 = 1 And Outward Unit normal vector = j So, The integral becomes ∬S-yj + 3zk·dS = ∬S2-yj + 3zk·jdS2 = ∬S2-yj + 3zk·jdS2 = ∬S2-ydS2 = ∬S2-16·1 dS2 = ∬S2-16 dS2⇒∬S-yj + 3zk·dS = ∬S2-16 dS2 .........2 Now, For S2 The limits of t are: 0 ≤ t ≤ 2π The limits of s are: 0 ≤ s ≤ 4 So, Equation (1) becomes ∬S2-16 dS2 = ∫02π∫04-16s dsdt = -∫02π∫0416s dsdt⇒∬S2-16 dS2 = -∫02π∫0416s dsdt Therefore, we get ∬S-yj + 3zk·dS = ∫02π∫04s2cos2t + 7s2sin2ts dsdt + ∫02π∫04-16s dsdtor∬S-yj + 3zk·dS = ∫t1t2∫s1s2s2cos2t + 7s2sin2ts dsdt + ∫t1t2∫s1s2-16s dsdtwheret1 = 0t2 = 2πs1 = 0s2 = 4Now, ∫02π∫04s2cos2t + 7s2sin2ts dsdt = ∫02π∫04s3cos2t + 7sin2t dsdt = ∫02π∫04s31 - sin2t + 7sin2t dsdt = ∫02π∫04s31 + 6sin2t dsdt = ∫02πs44041 + 6sin2t dt = 14∫02π44 - 0 1 + 6sin2t dt = 64∫02π1 + 6sin2t dt = 644t - 3 cost sint02π = 648π = 512π⇒∫02π∫04s2cos2t + 7s2sin2ts dsdt = 512π and -∫02π∫0416s dsdt = -∫02π16s2204 dt = -8∫02π42 - 0 dt = -128∫02π dt = -128t02π = -1282π - 0 = -256π⇒-∫02π∫0416s dsdt = -256π Therefore, we get ∬S-yj + 3zk·dS = ∫02π∫04s2cos2t + 7s2sin2ts dsdt + ∫02π∫04-16s dsdt = 512π + - 256π = 256πi.e., we get⇒∬S-yj + 3zk·dS = 512π + - 256π