Math 212. Practice Problems for the Midterm 3

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1 Math 1 Practice Problems for the Midterm 3 Ivan Matic 1. Evaluate the surface integral x + y + z)ds, where is the part of the paraboloid z 7 x y that lies above the xy-plane.. Let γ be the curve in the xy plane representing the graph of the function f x) x 3 + x for x 3. Let be the surface of revolution obtained by rotation of the curve γ around the y axis. Let F be the vector field given by F y,x + y,z. Evaluate the integral F n ds, where n is the unit normal vector to the surface that forms an angle smaller than 9 with the y-axis. 3. If F 3 i + y j, and if the surface is given by its parametrization R s,t) 9coss,9sins,t for s π and t 1, evaluate F n ds, where n is the unit normal vector that is pointing away from z axis at any point of the surface. 4. Find the area of the part of the sphere x + y + z 1 that lies above the xy plane and within the cylinder x + y ).. Evaluate the integral y dx + xydy + zdz where is the intersection of the paraboloids z x + 4y and y x 4z oriented counter-clockwise when viewed from the origin. 6. Evaluate the integral F ds, where is the curve with the parametrization r t) cost,sint,sin t for t π and x F is the vector field defined by F y, y x, z z. 1

2 7. onsider the surfaces S 1, S, and S 3, where S 1 is the surface generated by rotating the curve about the z-axis, S is the disk and S 3 is the disk y 1 + z 4, 1 z, x + y 4, z 1, x + y 89, z. Let denote the union of surfaces S 1, S, and S 3. Let E denote the solid whose boundary is. a) Plot the above surfaces. Parametrize S 1 and calculate S 1 f ds, where f b) Find the volume of E. z. 1+z 4 8. Let S 1, S, S 3, and be as in the previous problem. Orient these surfaces positively outward). a) Find the flux of e x +y y ) i + sinπz) 3 e x y j ) x + k across. b) Find the flux of accross S 1 S 1 is oriented with normal vector pointing outside of ). 9. Let S 1, S, S 3,, and be as in Problems 1 and. Orient these surfaces positively outward). a) Find the flux of across S 1 S 1 is oriented with normal vector pointing outside of ). b) Find the flux of across. F x + y + z ) 3 x i + y j + z k ) 1. onsider the polar curve r sinθ), θ π. a) Sketch the curve. b) Find its length. Do not evaluate the integral.) c) Find the area it encloses. 11. Find the circulation around an ellipse given by the intersection of the plane x + 6y 3z 6 with the cylinder x + y 1, for the vector field F y, y,3z. he ellipse is oriented counterclockwise for the person standing at the origin. 1. Assume that is a solid with boundary S. Assume that u and v are two times continuously differentiable functions such that v on the set S and u satisfies u xx + u yy + u zz f x,y,z). Prove that u vdxdydz f vdxdydz.

3 Solutions 1. Evaluate the surface integral x + y + z)ds, where is the part of the paraboloid z 7 x y that lies above the xy-plane. Solution. he parametrization of the part of the paraboloid that is above the xy-plane is: x r cosθ y r sinθ z 7 r, where r and θ are real numbers such that θ π and r 7. he parametrization can be written in the vector form as R r,θ) r cosθ,r sinθ,7 r, θ π, r 7. We have that ds R r R θ drdθ hence we need to calculate R r R θ. herefore he integral now becomes R r R θ cosθ,sinθ, r r sinθ,r cosθ, i j k det cos θ sin θ r r sinθ r cosθ r cosθ i + r sinθ j + r k r r cosθ,r sinθ,1. ds r 4r cos θ + 4r sin θ + 1drdθ r 1 + 4r drdθ. x + y + z)ds 7 14 r ) r 1 + 4r drdθ. We will evaluate the inner integral using the substitution u 1 + 4r. hen we have du 8r dr and as r goes from to 7, the variable u goes from 1 to 9. herefore the integral becomes: 7 14 r ) r 1 + 4r dr u 1 4 ) udu udu u 3 du u 3 u9 1 3 u u u9 u1

4 We are now ready to evaluate the original integral: 1dS dθ π.. Let γ be the curve in the xy plane representing the graph of the function f x) x 3 + x for x 3. Let be the surface of revolution obtained by rotation of the curve γ around the y axis. Let F be the vector field given by F y,x + y,z. Evaluate the integral F n ds, where n is the unit normal vector to the surface that forms an angle smaller than 9 with the y-axis. Solution. Assume that Mx,y,z) is a point on the surface of revolution, and let ˆMx,,z) be its projection to the xz plane. Let us denote by δm) its distance from the y-axis. Since the surface is obtained by rotating the curve around the y-axis, we must have y f δm)). We will parametrize the surface using parameters δ and θ, where δ is a distance of a point from the y axis, and θ is the angle that the line ˆMO forms with the y-axis. he parametrization of the surface is x δ cosθ y δ 3 + δ z δ sinθ, where δ and θ are real numbers such that θ π and δ 3. he parametrization can be written in the vector form as R δ,θ) δ cosθ,δ 3 + δ,d sinθ, θ π, δ 3. We have that n ds ± R δ R θ dδdθ where the sign will be chosen so that the obtained vector forms an angle smaller than 9 with the y axis. Now we need to calculate R δ R θ. R δ R θ cosθ,3δ + 4δ,sinθ δ sinθ,,δ cosθ i j k det cosθ 3δ + 4δ sinθ δ sinθ δ cosθ δ 3δ + 4)cosθ i δ sinθ j + δ 3δ + 4)sinθ k δ 3δ + 4)cosθ, 1,3δ + 4)sinθ. he obtained vector R δ R θ has a negative dot product with the vector,1, corresponding to the y axis. hat means that the angle it forms with the y-axis is greater than 9, therefore the sign should be chosen and we get: n ds δ 3δ + 4)cosθ,1, 3δ + 4)sinθ dδdθ. 4

5 he integral now becomes: π F n ds δ 3 + δ ),δ cosθ + δ 3 + δ,δ sinθ 3δ + 4)cosθ,1, 3δ + 4)sinθ δ dδdθ I 1 + I I 3 where I 1 δ 3 + δ ) δ 3δ + 4)cosθ dδdθ I δ cosθ + δ 3 + δ ) δ dδdθ I 3 3δ + 4)δ 3 sin θ dδdθ he first integral can be evaluated in the following way: I 1 Let us now evaluate the second integral: I δ 3 + δ ) δ 3δ + 4)cosθ dδdθ δ 3 + δ ) π δ 3δ + 4)dδ cosθ dθ. δ 3 dδ δ cosθ + δ 3 + δ ) δ dδdθ δ 3 cosθ dδdθ + cosθ dθ + π + π π 3 37 π. It remains to calculate the third integral: I 3 14 π 34 3δ + 4)δ 3 sin θ dδdθ 1 cosθ) dθ We are now ready to evaluate the original integral: F n ds I1 + I I 3 37 π 13π. δ + ) δ 4 dδdθ δ dδ + 4π δ 4 dδ sin θ dθ 3δ 4 + 4δ 3) dδ π π 34 3δ + 4) δ 3 dδ ) π π ) 34 13π

6 3. If F 3 i + y j, and if the surface is given by its parametrization R s,t) 9coss,9sins,t for s π and t 1, evaluate F n ds, where n is the unit normal vector that is pointing away from z axis at any point of the surface. Solution. We have that n ds ± R s R t dsdt where the sign will be chosen so that the obtained vector points away from z axis. R s R t 9sins,9coss,,,1 i j k det 9sins 9coss 1 9coss,9sins,. he obtained vector R s R t points away from z axis at any point because the x and y components of the vector at x,y,z ) are precisely equal to x and y. hus the sign + should be chosen in place of ±. he integral now becomes: F n ds 1 3,9sins, 9coss,9sins, dtds I 1 + I where 1 I 1 7cossdtds I 1 81sin sdtds. We evaluate the integrals I 1 and I in the following way: I 1 I 7cossds. 81sin sds 4 1 coss)) ds 81π. We are now ready to evaluate the original integral: F n ds I1 + I 81π. 6

7 4. Find the area of the part of the sphere x + y + z 1 that lies above the xy plane and within the cylinder x + y ). Solution. We need to evaluate the integral 1dS, where is the described surface. Let us first parametrize the surface. Our parameters will be r and θ and the parametrization of the sphere above the xy-plane is: x r cosθ y r sinθ z 1 r, where r and θ are real numbers such that r 1 and θ π. he equation of the cylinder can be rewritten as x + y 1y, or, equivalently r 1r sinθ. he point x,y,z) is within the cylinder if and only if the corresponding r and θ satisfy θ π and r 1sinθ. hus the parametrization of is: R r,θ) r cosθ,r sinθ, 1 r for θ π and r 1sinθ. We have that ds, R r R θ drdθ hence we need to calculate R r R θ. R r r R θ cosθ,sinθ, r sinθ,r cosθ, 1 r i j k det cosθ sinθ r 1 r r sinθ r cosθ herefore ds r cosθ i + 1 r r 1 r r sinθ 1 r j + r k r cosθ,r sinθ, 1 r. r r cos θ + r sin θ + 1 r 1r drdθ drdθ. 1 r 1 r he integral now becomes 1dS π 1sinθ 1r 1 r drdθ. We will evaluate the inner integral using the substitution u 1 r. hen we have du r dr and as r goes from to 1sinθ, the variable θ would travel from 1 to 1 1sin θ 1cos θ. herefore the integral becomes: 1sinθ 1r 1 r dr 1cos θ 1 du u 1u 1 u1 u1cos θ 1 1 cosθ cos θ u 1 du

8 We are now ready to evaluate the original integral: 1dS π 1 1 cosθ ) dθ π 1 π. π π cosθ dθ 1 cosθ)dθ π. Evaluate the integral y dx + xydy + zdz where is the intersection of the paraboloids z x + 4y and y x 4z oriented counter-clockwise when viewed from the origin. Solution. We can use Stokes heorem to evaluate this integral. he vector field is F y,xy,z and its curl is: i j k F det x y z,,y y. y xy z Since the curve is closed, we can denote by S the surface whose boundary is for example the portion of the first paraboloid). he integral is equal to. 6. Evaluate the integral F ds, where is the curve with the parametrization r t) cost,sint,sin t for t π and x F is the vector field defined by F y, y x, z z. Solution. Let us first write F + H, where x 4 +, y 4 + 3, z z and H y, x,z. It is easy to find that,,. Hence according to Stokes heorem we have that F d r d r + H d r d S + H d r H d r, S where S is any surface with boundary. For the last integral we use the parametrization of and obtain: π H d r sint, cost,sin t sint,cost, 1 cos t dt + 1 ) sint dt 4π. 8

9 he solutions to the problems 7 11 can be located in the document with title Practice final on the course web page for Math 13: Multivariable alculus, Fall Assume that is a solid with boundary S. Assume that u and v are two times continuously differentiable functions such that v on the set S and u satisfies u xx + u yy + u zz f x,y,z). Prove that u vdxdydz f vdxdydz. Solution. onsider the vector field F v u vu x,vu y,vu z. hen we have F x vu x) + y vu y) + z vu z) vu xx + u yy + u zz ) + u x v x + u y v y + u z v z f v + u v. Since v on S we have u v d S. Using the Divergence heorem we obtain: S S u v d S he last equality implies the desired result. u v) dv f vdv + u vdv. 9

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