Engineering Mathematics 233 Solutions: Double and triple integrals

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1 Engineering Mathematics s: Double and triple integrals Double Integrals. Sketch the region in the -plane bounded b the curves and, and find its area. The region is bounded b the parabola and the straight line. The points of intersection of the two curves are given b,. This gives the two points A, and,. A The region is a Tpe II region, and can be described b area :. da d 8 6. / 6 d d. Evaluate the integral b interchanging the order of integration. + d d

2 The region of integration is the Tpe II region :. We have. from the drawing above, we can rewrite the region as the Tpe I region :. + d d + + d d d Find the volume of the region of bounded b the paraboloid z + and b the planes z, a, a, a and a.

3 Let S be the D region bounded b the paraboloid and the planes. z a -a -a a region volumes where is the projection of S in the -plane, i.e. volumes : a a a + da + da, a a a a. a a a + d d a + a a 8a. a a d a + d d + a + a d a. Evaluate + da, where is the region of the plane given b + a. The region and the integrand + are best described with polar coordinates r, θ. In those coordinates, the region, which is the region inside the circle + a, becomes : + da r a θ π. π a π a π a r r dr dθ r dr dθ dθ πa.

4 5. Evaluate e + da, where is the region of. above. Using polar coordinates again, we write e + da π a e r π π e r a dθ r dr dθ e a dθ π e a. 6. Evaluate the integral e d d. Hint: First reverse the order of integration. If we tr to evaluate the integral as written above, then the first step is to compute the indefinite integral e d. ut e does not have an indefinite integral that can be written in terms of elementar functions. we will fist reverse the order of integration. The region of integration is the Tpe II region :. can also be described as the Tpe I region This gives :. e d d e e e e d d d d e.

5 7. Find the volume of the tetrahedron bounded b the coordinate aes and the plane z. We have to find the volume of the tetrahedron S bounded b the plane z z 6 and the coordinate aes. This is the portion of the plane in the first octant, as one can see from Picture. z egion plane z line + 6 we have volumes 6 where is the projection of the tetrahedron in the -plane. is the Tpe I region see Picture da, Finall, this gives :. 6 volumes /6 6 d d 6 + d +. d d 5

6 8. Evaluate the integral + d d. Hint: Use polar coordinates. The region of integration is the Tpe II region : We have +. varies between the straight line and the circle +. The region is P, In polar coordinates, the region is : r θ α, where α is the angle made b the straight line. The straight line and the cicle meet at the points + ±. The intersection point in the first quadrant is then P, cos α, sin α. α arcsin π 6. Finall, the integrand + is r in polar coordinates. This gives + d d π 6 π/6 π/6 r r r dr dθ dθ 8 π Find the volume below the surface z +, above the plane z, and inside the clinder +. 6

7 Completing the squares, we rewrite the equation of the clinder as The base of the clinder is then the circle of radius centered at,. we have to find the volume of the D region: z z + volume region From the picture above, we write V + da, where is the projection of the D region in the plane, i.e. the circle +. Using polar coordinates, this gives V r r dr dθ. In polar coordinates r cos θ and r sin θ, the circle writes as + r r sin θ r sin θ, and is the region : θ π r sin θ. V π sin θ π π 8 r dr dθ sin θ dθ π. π r sin θ dθ θ 8 sin θ + sin θ π Triple Integrals. Evaluate z + z dz d d. 7

8 + z dz d d z d d z + + d d d d 8 d 8 d 7 8 d Find the mass of the D region given b + + z,,, z, if the densit is equal to z. We have mass z dv, and the region is the portion of the sphere of radius in the first octant. z sphere + + z + 8

9 D region D region can be described as z, for all,, where is the projection of in the -plane. Describing as a Tpe I region, this gives mass : z. z dz d d 5 z d d d d d d + d d d. Find the volume of the region bounded b the paraboloid z and the -plane. 9

10 We have volume dv. z,, z - - can be described as z, for all,, where is the projection of in the -plane. is the interior of the circle +. In polar coordinates, the region is : θ π r, and in clindrical coordinates, the region is volume : z r π π θ π r. π r π r dz dr dθ r r dr dθ r r dr π 8π. r r. Find the center of gravit of the region in., assuming constant densit σ. smmetr,. Also, as the densit is d,, z σ, d,, zz dv σ z d,, z dv σ z dv volume z dv. 8π z dv dv

11 Using the description of the region in clindrical coordinates of., we get z dv π r π z r dz dr dθ r r dr dθ π 6r 8r + r 5 dr π 8r r + r6 6 π. z 8π π.. Evaluate + + z dv, where is the region bounded b the plane z and the cone z +. We will use the spherical coordinates z φ ρ θ to describe the region. In those coordinates, ρ sin φ cos θ ρ sin φ sin θ z ρ cos φ. the cone z + writes as ρ cos φ ρ sin φ sin φ cos φ tan φ φ π, the plane z as ρ cos φ ρ cos φ,

12 and the region can be described as : φ π φ π ρ cos φ. z z + Finall, in spherical coordinates, + + z ρ. + + z dv π π π π/ / cos φ π/ π/ sin φ π/ / cos φ ρ sin φ ρ ρ sin φ dρ dφ dθ / cos φ π sin φ cos φ dφ 8π cos φ π/ 8π 6 ρ dρ dφ dφ 7π. 5. Evaluate + + z / dv, where is the region bounded b the spheres + + z a and + + z b, where a > b >. Using spherical coordinates, the region between the spheres can be described as : θ π φ π a ρ b.

13 z + + z / dv π π b a π b ρ ρ sin φ dρ dφ dθ π sin φ dρ dφ a ρ π π ln ρ b sin φ dφ a b π π ln sin φ dφ a b π ln cos φ π π ln a b. a 6. Find the volume of the region bounded above b the sphere + + z a and below b the plane z b, where a > b >. The region can be described as the set of,, z such that b z a

14 for all, in the plane region bounded b the circle + + b a + a b.,, a z b sphere + + z a The region is best described in clindrical coordinates, and this gives : b z a r r a b θ π. volume π π dv π a b a r π a b a b π b r dz dr dθ a r br dr dθ a r r br dr a r / / a b br b / a / b a b a b π a b + b a π a b + b Sketch the region whose volume is given b the triple integral / 6/ ewrite the triple integral using the order of integration dv d d dz. From the triple integral, the region is described b z 6, dz d d.

15 6 i.e. z varies between the planes z and z the region described b z. Furthermore,, are in :, which is the projection of the plane z in the -plane. z,,,,,, region We now use the ordre of integration dv d d dz. The region be can be described as 6 for all, z in the region which is the projection of in the z-plane. can be described as This gives : z z. / 6/ dz d d z/ z/6 d d dz. 8. Evaluate + dv, where is the region ling above the -plane, and below cone z +. The region be can be described as z, for all,, where is the projection of in the -plane. is the region inside the curve + + 6, 5

16 which is a circle of radius. z z + We then use clindrical coordinates to describe the region. This gives : z r r θ π. + dv π r π π π 6 r r dz dr dθ r r dr dθ r r dr r r π dθ dθ 6 π. 9. Evaluate the integral dz d d. Hint: First convert to clindrical coordinates. The region described b the integral is the region given b z 6, i.e. bounded below b the plane z and above b the paraboloid z 6, for all,. For the region, we have 6, i.e. varies between the straight line and the top part of the circle + 6. Similarl, varies between 6

17 the straight lines and. is the portion of the circle + 6 in the first quadrant. z,,6 z D region D region The region is best described in clindrical coordinates as 6 6 : z 6 r θ π/ r. + dz d d π π/ 6 r r 6 r dr π 6r r dr π 6r r5 5 π 8 π 5 5. r r dz dr dθ. Evaluate + + z dv, where is the region above the -plane bounded b the cone z + and b the sphere + + z. In spherical coordinates ρ, θ, φ, the sphere is + + z ρ ρ, and the cone is z + ρ cos φ ρ sin φ tan φ tan φ ± φ π 6 or π 6. 7

18 The part of the cone above the -plane corresponds to φ π 6. z sphere + + z top part of cone z + in spherical coordinates, the region is : ρ φ π 6 θ π. + + z dv π π/6 π π/6 π π/6 π π ρ ρ sin φ dρ dφ dθ ρ sin φ dφ dθ sin φ dφ dθ π/6 cos φ dθ dθ π. 8

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