Green's, Stokes', and the divergence theorem #2 - Questions and Answers

Question 1 Apply Green's theorem to find the outward flux for the field 𝐹 ( 𝑥 , 𝑦 ) = tan ⁡ − 1 ( 𝑦 𝑥 ) 𝑖 + ln ⁡ ( 𝑥 2 + 𝑦 2 ) 𝑗 . F(x,y)=tan −1 ( x y ​ )i+ln(x 2 +y 2 )j. Answer Step 1 We are given that 𝐹 ( 𝑥 , 𝑦 ) = tan ⁡ − 1 ( 𝑦 𝑥 ) 𝑖 + ln ⁡ ( 𝑥 2 + 𝑦 2 ) 𝑗 F(x,y)=tan −1 ( x y ​ )i+ln(x 2 +y 2 )j and the curve 𝐶 C defined as the boundary of the given region. We will be using Green's theorem. Question 2 Use Green's Theorem to evaluate the line integral. Orient the curve counterclockwise. ∮ 𝐶 𝐹 ⋅ 𝑑 𝑟 ,  where  𝐹 ( 𝑥 , 𝑦 ) = ⟨ 𝑥 2 , 𝑥 2 ⟩ ,   𝐶  consists of the arcs  𝑦 = 𝑥 2  and  𝑦 = 8 𝑥  for  0 ≤ 𝑥 ≤ 8. ∮ C ​ F⋅dr, where F(x,y)=⟨x 2 ,x 2 ⟩,C consists of the arcs y=x 2  and y=8x for 0≤x≤8. Answer Step 1 We have given the line integral with values, 𝐹 ( 𝑥 , 𝑦 ) = ( 𝑥 2 , 𝑥 2 ) and arcs 𝑦 = 𝑥 2 , 𝑦 = 8 𝑥 . F(x,y)=(x 2 ,x 2 )andarcsy=x 2 ,y=8x. Step 2 We know Green's theorem and can proceed using the theorem. Question 3 Randomly drawing three green skittles in a row from a bag that contains eight green skittles out of 25 skittles total. Answer Step 1 Given that a bag contains 25 skittles in which eight are green. Three green skittles are drawn from the bag in a row, and the skittles are drawn eaten immediately. Therefore, Total number of skittles = 25 and total number of green skittles = 8. Step 2 Probability that the 1st skittle drawn from the bag is green will be Number of green skittles/Total number of skittles = 8 25 . Number of green skittles/Total number of skittles= 25 8 ​ . The 1st drawn skittle has been eaten; therefore, there are 24 skittles left in which 7 skittles are green. Probability that the 2nd skittle drawn from the bag is green will be Number of green skittles after 1st draw/Total number of skittles after 1st draw = 7 24 . Number of green skittles after 1st draw/Total number of skittles after 1st draw= 24 7 ​ . The 2nd drawn skittle eaten immediately; therefore, there are 23 skittles left in which 6 skittles are green. Probability that the 3rd skittle drawn from the bag is green will be Number of green skittles after 2nd draw/Total number of skittles after 2nd draw = 6 23 . Number of green skittles after 2nd draw/Total number of skittles after 2nd draw= 23 6 ​ . Step 3 Probability ( 𝑝 p) that the three green skittles are drawn in a row can be obtained as follows: 𝑝 = Probability that the 1st skittle drawn from the bag is green × Probability that the 2nd skittle drawn from the bag is green after 1st draw × Probability that the 3rd skittle drawn from the bag is green after 2nd draw . p=Probability that the 1st skittle drawn from the bag is green×Probability that the 2nd skittle drawn from the bag is green after 1st draw×Probability that the 3rd skittle drawn from the bag is green after 2nd draw. 𝑝 = 8 25 × 7 24 × 6 23 . p= 25 8 ​ × 24 7 ​ × 23 6 ​ . 𝑝 = 0.0243. p=0.0243. Therefore, the probability that the three green skittles are drawn in a row is approximately 0.0243 0.0243. Question 4 Use the Divergence Theorem to calculate the surface integral 𝐹 ⋅ 𝑑 𝑆 F⋅dS, i.e., calculate the flux of 𝐹 F across 𝑆 S. 𝐹 ( 𝑥 , 𝑦 , 𝑧 ) = 𝑥 3 + cos ⁡ 𝑦   𝑖 + 𝑦 3 sin ⁡ 𝑧   𝑗 + 𝑧 3   𝑘 . F(x,y,z)=x 3 +cosyi+y 3 sinzj+z 3 k. 𝑆 S is the surface of the solid bounded by the cylinder 𝑦 2 + 𝑧 2 = 4 y 2 +z 2 =4 and the planes 𝑥 = 0 x=0 and 𝑥 = 4 x=4. Answer To solve this problem, we will use the Divergence Theorem, which relates the flux of a vector field across a closed surface to the divergence of the vector field within the enclosed volume. First, let’s find the divergence of 𝐹 F: ∇ ⋅ 𝐹 = ∂ ∂ 𝑥 ( 𝑥 3 + cos ⁡ 𝑦 ) + ∂ ∂ 𝑦 ( 𝑦 3 sin ⁡ 𝑧 ) + ∂ ∂ 𝑧 ( 𝑧 3 ) . ∇⋅F= ∂x ∂ ​ (x 3 +cosy)+ ∂y ∂ ​ (y 3 sinz)+ ∂z ∂ ​ (z 3 ). = 3 𝑥 2 − sin ⁡ 𝑦 + 3 𝑧 2 . =3x 2 −siny+3z 2 . Now, we need to find the surface 𝑆 S, which is the surface of the solid bounded by the cylinder 𝑦 2 + 𝑧 2 = 4 y 2 +z 2 =4 and the planes 𝑥 = 0 x=0 and 𝑥 = 4 x=4. This surface consists of three parts: the top and bottom surfaces of the cylinder (which are circles), and the curved surface of the cylinder. We can calculate the flux across each of these surfaces separately and add them up to get the total flux across 𝑆 S. To find 𝑑 𝐴 dA, we can use polar coordinates: 𝑥 = 4 cos ⁡ 𝜃 ,   𝑦 = 2 sin ⁡ 𝜃 ,   and  𝑑 𝐴 = 𝑟   𝑑 𝑟 𝑑 𝜃 , x=4cosθ,y=2sinθ,and dA=rdrdθ, where 𝑟 r is the radial distance from the origin. Substituting these expressions into the integral, we get: ∬ 𝑆 𝐹 ⋅ 𝑑 𝑆 = ∬ 𝐷 ( 4 3 cos ⁡ 3 𝜃 + cos ⁡ ( 2 sin ⁡ 𝜃 ) ) 𝑟 𝑑 𝑟 𝑑 𝜃 . ∬ S ​ F⋅dS=∬ D ​ (4 3 cos 3 θ+cos(2sinθ))rdrdθ. = 32 𝜋 3 . = 3 32π ​ . Therefore, the flux of 𝐹 F across 𝑆 S is 32 𝜋 3 3 32π ​ .