Math 263 Assignment 7 SOLUTIONS

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1 Problems to turn in: Math 6 Assignment 7 SOLUTIONS In each case sketch the region and then compute the volume of the solid region. a The ice-cream cone region which is bounded above by the hemisphere z a x y and below by the cone z x + y. Solution. In spherical coordinates, V V π π/4 a π/4 ρ sinφdρdφdθ π sin φρ / ] a dφ π cos φ] π/4 a / πa cos π/4 + cos πa or in cylindrical coordinates, π a/ a r a r / r π a a / π r rdzdrdθ π ] a/ a/ r a r r dr π a a / / + a / a/ + πa. b The region bounded by z x + y and z 4 y. Solution. The parabolic cylinder z 4 y comprises the top of the surface considered in terms of z and the paraboloid z x + y is the bottom surface in terms of z. To determine the region of the xy-plane which the region bounded by these two surfaces lies over, we intersect the two surfaces, in this case we can set them equal to each other. We see that x + y 4 y if and only if x + 4y 4 if and only if x/ + y. We will set up and compute the integral of this volume in rectangular coordinates using a table of integrals to compute the anti-derivative 4 x / dx. V 4 x / 4 y 4 x / x +y dzdydx 4 x / 4 x / 4 x y 4/y ] 4 x / 4 x / dx x 4 / dx 4sin sin 4π 4 x 4y dydx 4 x / 4 x / dx 6 ] x x 4 x sin x/ 8 Another way to compute this integral would be to make a substitution x u, so dx du and we would be integrate over a circle of radius in u,y, which we will call R whereas the ellipse will be called R. This makes everything much simpler. Lets see what happens.

2 V 4 y dz R x +y π da 4 4r rdrdθ R π 4 x 4y dxdy dθ 8r 8r dr π 4r r 4] 4 4u 4y dudy er π4 4π c A sphere with a cylindrical hole bored through its centre. Specifically, the region inside the sphere x + y + z 9 and outside the cylinder x + y 4. Solution. A sphere of radius has volume V S 6π. Let V denote the volume inside the given sphere and the given cylinder simultaneously. The the volume we want, V V S V. Let s compute V using cylindrical coordinates. V π 9 r / π π rdzdrdθ 9 r / rdrdθ 9 r / ] 9 r / 4π 9/ / 6π 4π/ ; hence V V S V 4π/. Switch these integrals to spherical coordinates and compute: Solution. I is an integral over the top half of a solid sphere of radius, centred at the origin. I 9 x 9 x π/ π φ θ 9 x y z x + y + z dzdydx ρ π/ π sin φcos φdφ sin φ ] π/ ρcos φ] ρ ρ sin φdρdθdφ π 4π dθ ρ 4 dρ I is a solid region contained within x >,y >,z >. The solid is above the cone z x + y and below the sphere x + y + z 8. To check this, note that the cone meets the sphere at the height where z + z 8, z, and the ring where they intersect is x + y 9. The angle of the point of the bottom of the cone is φ π/4. Putting this together, we have

3 I 9 y 8 x y x + y + z dzdydx x +y π/4 π/ 8 φ θ ρ ρ ]ρ sin φdρdθdφ π/4 π/ 8 sin φdφ dθ ρ 4 dρ φ cos φ] π/4 θ π ρ 486π alculate the moment of inertia of a circular pipe of outer radius a, inner radius b, length L and uniform density R, rotating about its centre axis. From your answer, let b and derive the formula for a solid cylinder too. Solution. Line the cylinder up along the z-direction and then the integral we need is easy to do in cylindrical coordinates: π L a x + y RdV R r rdrdzdθ πlra4 b 4. Letting b, we obtain the moment of inertia of a solid cylinder, /πlra 4. b 4 Find the gradient vector field of fx,y x + y and gx,y x y. In each case, plot the gradient vector field and the contour plot of the function, on the same diagram. f x y x + y,, g x, x + y ompute fx,y,zds for the following curves and functions. a : rt cos t,sin t for t π/ and fx,y + y/. Solution. First, ds r t dt 9cos t sin t + 9sin t cos t dt 9cos t sin tdt. Now we are in a position to compute the line integral.

4 + y/ds π/ u + sin t9cos t sin tdt 9u + 9u 4 du, where u sint 4u + 8u ] π/ 9sin t + 9sin 4 tcos tdt b : rt t /,t / for t and fx,y x + y. Solution. Again we start by computing ds r t dt t + t dt. Then x + y ds 8 8 t / + t / t + t dt 4 u u 7/ u udu + 8 u u / u / + u / du + 8 t 4 + t tdt + 9 u udu, where u + t 8 u/ + u/ + u9/ 8 u7/ + u/ u/ 7 u 9/ 8 u7/ 84 u/ + u/ 8 ] u 7/ u / + u / u / du ] t 6 + t tdt 9/ /8 7/ /84 / / + / /8 /8 /84 / + /8 c : rt,,t for t and fx,y,z e z. Solution. e z ds e t + + t dt te t dt te t e t ] Note that we had to integrate by parts to anti-differentiate te t. You let u t and dv e t. 6 Determine whether or not the following vector fields are conservative. In the cases where F is conservative, find a function ϕ such that Fx, y, z ϕx, y, z. a F xy + z i + x + yzj + y + xzk. Solution. We first test to determine whether or not F might be conservative. Letting F xy + z, F x + yz, and F y + xy as usual, it is easy to verify that F / y F / x, F / z F / x, and F / z F / y. There are many ways to find a function ϕx,y,z such that ϕ F, which is what we need to find. Here is one method. We will take antiderivatives of F with respect to x, F with respect to y, and F with respect to z respectively and then compare the results. ϕx,y,z xy + z dx x y + xz + y,z ϕx,y,z x + yzdy x y + y z + x,z ϕx,y,z y + xzdz y z + xz + x,y 4

5 It is very important that y,z is function of y and z and not just a constant, since we are undoing a partial derivative where we considered y and z as constants similarly for x,z and x,y. If we examine the three versions of ϕx,y,z we see that each version has at least one term in common. Therefore, we might try ϕx,y,z x y + y z + xz, which turns out to work in this case. b F lnxyi + x y j + yk. Solution. Note that F is only defined for x,y > or x,y < and F lnxy, F x/y, and F y have continuous partials in these regions of the plane. Further, if F ϕ, and hence F is conservative, then the mixed second partials of ϕ must be equal. But since F / z and F / y, no such ϕ could exist with ϕ lnxyi + x y j + yk. c F e x cos yi + e x sin yj + zk. Solution. By inspection, it is easy to see that ϕx,y,z z + e x cos y is a potential function for F. Otherwise, one could use a method similar to a.