1.(6pts) Which integral computes the area of the quarter-disc of radius a centered at the origin in the first quadrant? rdr d

Transcription

1 .(6pts) Which integral computes the area of the quarter-disc of radius a centered at the origin in the first quadrant? (a) / Z a rdr d (b) / Z a rdr d (c) Z a dr d (d) / Z a dr d (e) / Z a a rdr d.(6pts) Find the moment about the y axis of a triangle with vertices (, ), (, ), (, ) and density distribution (x, y) =x. (a) / (b) /5 (c) /5 (d) /6 (e) / ZZZ.(6pts) ompute the triple integral x =( y), z =andz =+y + x. V dv over the region V bounded by x = y( y + x y), (a) (b) 8 (c) 6 (d) 5 9 (e) 8.(6pts) onsider the change of variables x =u+v, y = u v. Find the Jacobian (a) v (b) v (c) x + y (d) y (e) u + v). 5.(6pts) Use the change of variables x =u, y = u + v to rewrite the following integral in terms of u and v: Z Z x/+ x/ y dy dx (a) Z Z (u + v) du dv (b) Z Z (u + v) du dv (c) Z Z (u + v) du dv (d) Z Z y du dv (e) Z Z y du dv

2 Z 6.(6pts) Evaluate the line integral starting at (, ). xy ds where is the right half of the circle x + y =6 (a) (b) (c) (d) / (e) / 7.(6pts) Find the function p such that rp = y sin(z), xy sin(z),xy cos(z) and p(,, ) =. What is p,,? (a) (b) 8 (c) (d) (e) I 8.(6pts) Evaluate y e sin(x) cos(x ) dx + 5x +arctan(y ) dy where is the circle x + y =withcounter-clockwiseorientation. (a) 8 (b) (c) (d) (e) 9.(6pts) Let E be the solid determined by x + y 6 z,x + y + z 6 9andz >. Which of the following expresses ZZZ p x + y + z dv as iterated integrals correctly? E Hints: Depending on how you do your calculations one of the following formulas may be p! helpful: arccos = / =arctan(). (a) Z sin d d d (b) Z d d d (c) Z sin d d d (d) Z cos sin d d(e) d Z cos sin d d d.(pts) ompute the mass of the disc D centered at the origin with radius and density (x, y) =+y.

3 .(pts) A tetrahedron has vertices at the points (,, ), (,, ), (,, ) and (,, ). ompute the mass of the tetrahedron if the density function is (x, y, z) =cos z..(pts) Let R denote the parallelogram in the xy-plane bounded by the lines y = x, y = x, y =x and y =x +. omputetheintegral ZZ (x + y) da using a suitable change of variables. R

4 .(pts) (a)(pts) Let be a curve which is the boundary of a bounded region D in the plane. Suppose the density of D, µ(y) isafunctionofy. I Showthatthemomentaboutthex axis of this region with density µ can be found by xyµ(y) dy where is oriented as in Green s Theorem. (b)(pts) Here is a graph of the parametric curve r(t) = t t, t + t, 6 t 6. Is the orientation given by this parametrization the same as the one required by Green s Theorem? (c) (6pts) Suppose the density of this region is. Show that the moment about the x-axis is Z t (t )(t +)dt

5 . Solution. By inspection. Bounds are r =..a, =.. /, integrand is da = rdr d.. Solution. The coordinates of the center of mass are x = M y /M,ȳ = M x /M,sowecompute each of these quantities. The integrals that appear are easy to compute, since they are just polynomials in x and y. ZZ M = ZZ M x = = Z ZZ M y = (x, y)da = y (x, y) da = Z Z x Z Z x x x + x dx = x (x, y) da = Z Z x dy dx = x Z yx dy dx = x x dy dx = x ( x) dx = Z x + x5 5 Z Z y x x dx = x y x = dx = Thus, x =(/6)/(/) = /5, ȳ =(/)/(/) = /5. Z (x x ) dx = x Z ( x) x dx x = ; + = ( 5 + 6) = ; (x x ) dx = x x 5 5 = 5 =

6 . Solution. The curves x = y( y), x =( y) intersect at y( y) =( y) or y =and y = y so y =. Thus ZZZ V y + x dv = = = Z Z y( y) Z +y+x ( y) Z Z y( y) Z Z = = ( y) y + z +y + z dz dx dy y + x dx dy = y( y) ( y) dy = Z Z / +6y y dy = y +y Z y( y) ( y) dx dy = y y ( y + y ) dy y = 5 =. Solution. The =( v) = v.

7 5. Solution. The domain of integration in terms of u, v is 6 u 6 and 6 v 6, since the equations y = x/ andy = x/+arev =andv =intermsofu, v. The Jacobian is So the integral v) Z Z = =. (u + v) du dv. 6. Solution. We can parametrize as hcost, sinti for apple t apple. Thends = p ( sint) +(cost) dt = dt. Thenthelineintegralbecomes cost sint dt =6 cos t sin tdt=sin (t) = = y sin(z) sop = xy =xy sin(z) soh(y, = xy cos(z)+g (z) =xy cos(z) so p(x, y, z) =xy sin(z). Since p(,, ) = this is the desired function. Thus p,, = sin =.

8 8. Solution. M =y e sin(x) cos(x )andn =5x @y ZZ =5 =so by Green s Theorem, y e sin(x) cos(x ) dx + 5x +arctan(y ) dy = da where D is the disk of radius so the answer is =8. D 9. Solution. We use spherical coordinates. Then the three inequalities defining the solid can be rewritten as 6 9 p 6 cos ) cos > cos > ) cos > This yields and 6 6 Therefore the iterated integrals should be Z sin d d d. Solution. We integrate straightforwardly to. Z Z D da = Z ( + rsin( ))rdrd which evaluates quite

9 . Solution. First find the equation of the plane: (i + j) (i + k) =i j k so x y z = so z = x y. ZZZ V cos z dv = Z Z x Z x Z Z x y cos z dz dy dx x = sin z y dy dx = Z Z x sin (x y) dy dx = Z x cos (x y) dx = Z cos x dx = 8 sin x = 8

10 . Solution. Let u = x + y and v =x y. ThentheregionR in the uv-plane is 6 u 6, 6 v 6. We can solve for x and y: x = (u + v) The Jacobian v) = Therefore the integral in terms of u, v is Z Z u du dv == y = (u Z u v). =. dv = 9 Z dv =. I I I. Solution. For part (a), by Green s Theorem xyµ dy = dx+xyµ dy = h,xyµi r da. Sinceµ is a function of I ZZ = yµ, so xyµ dy = yµ @x D which is the formula for the moment about the x axis for the region D with density µ. For part (b), the tangent vector to the curve at the point r(t) isr (t) = t, t +t. Pick any point, say t =. Thenatr() = h, i, r () = h, i. ForGreen stheoremyou need the orientation so that as you walk around the curve in the preferred direction to the region is on your left. Hence the preferred direction at h, i is <, > so the orientation given by the parametrization I is opposite Z to the one given by Green s Theorem. Z For part (c), we do xy dy = (t t)(t + t )(t +t) dt = t (t )(t + )(t +)dt = Z t (t )(t +)dt Z If you actually want to know the answer t (t 6 + t t ) dt = (t + t 8 t t 6 t ) dt = + t9 t 7 t 5 = Youwerenotaskedfor the answer so there was no credit for just doing the integral. Some credit would have been given for arguing that the integral is negative but the moment is positive so the sign must be andhencetheparametricorientationisoppositetothe Green s Theorem orientation. Z

11