MATH 251.504Practice Problems for Examination 2Fall 20061. Find the absolute minimum and maximum values of the function f (x, y) = xy − 6x2on the closed region bounded by the parabola y = 9x2− 1 and the x-axis.2. Calculate the double integral.(a)RRR3xex2dA, where R = [−1, 0] × [0, 1].(b)RRR5 cos(y2) dA, where R is the region bounded by the lines y = −x, x = 0, andy = −1.(c)RRRex2+y2tan−1(yx) dA, where R is the region described in polar coordinates by{(r, θ) : 0 ≤ θ ≤π4, 0 ≤ r ≤ θ}.(d)RRR(5x2y+2xy5) dA, where R is the region between the two parabolas y = x2+10and y = 2x2+ 6.3. Find the volume of the solid which lies between the paraboloid z = 6x2+ y2and thexy-plane and above the region {(x, y) : 0 ≤ y ≤ 1, y2≤ x ≤ y}.4. Find the center of mass of a lamina occupying the region {(x, y) : 0 ≤ x ≤ 5, 0 ≤ y ≤x2} with density function ρ(x, y) = x2.5. Determine whether the following is true or false: −2 ≤RRRx sin(exy2+ x5) dA ≤ 2,where R = [0, 1] × [−2, 0].6. ComputeR0−1R1−x(e−y2+ xy) dy dx.7. Find the volume of the solid bounded by the parabolic cylinder z = y2and the planesz = 4y, x = 1, and x = 2.8. Consider a solid object which occupies the space between the parabolic cylinders z = y2and z = y2+ 2 and over the region bounded by the lines x = 0, y = 0, and y = 1 − x.Suppose the object has density function ρ(x, y, z) = xyz. Find the moments of inertiaof the object about the x- and y-axes.1Solutions1. We have fx(x, y) = y − 12x and fy= x, and both of these are zero at the point(x, y) = (0, 0), which is on the boundary of the region. For the part of the boundary ofthe region that lies on the x-axis we have the function f(x, 0) = −6x2for −13≤ x ≤13,which has maximum value 0 at 0 and minimum value −23at13and −13. On the otherpart of the boundary we have the function g(x) = f(x, 9x2− 1) = 9x3− 6x2− x for−13≤ x ≤13. Then g0(x) = 27x2− 12x − 1, which is zero at x0=12−√25254. Sinceg00(x0) < 0 the value g(x0) ≈ 0.03754 is maximum for g, while g(13) = g(−13) = −23is the minimum. Thus for f the absolute minimum value is −23while the absolutemaximum value is approximately 0.03754.2. (a)Z10Z0−13xex2dx dy =Z10h32ex2ix=0x=−1dy =32(1 − e).(b)Z0−1Z−y05 cos(y2) dy =Z0−1h5x cos(y2)ix=−yx=0dy =Z0−1−5y cos(y2) dy= −52sin(y2)i0−1=52sin(1).(c)Zπ40Zθ0θrer2dr dθ =Zπ40h12θer2ir=θr=0dθ =Zπ4012θ(eθ2− 1) dθ=14eθ2−14θ2iπ40=14eπ216−π264−14.(d)Z2−2Zx2+102x2+6(5x2y + 2xy5) dy dx =Z2−2h52x2y2+13xy6iy=x2+10y=2x2+6dx=Z2−252x2((x2+ 10)2− (2x2+ 6)2)dx +Z2−213x((x2+ 10)6− (2x2+ 6)6)dx.Expand to evaluate the first integral, and for the second use substitution.23. The volume is given byV =Z10Zyy2(6x2+ y2) dx dy =Z10h2x3+ y2xix=yx=y2dy =Z10(3y3− 2y6− y4) dy=34y4−27y7−15y5i10=37140.4. The mass and moments arem =Z50Zx20x2dy dx =Z50x4dx =15x5i50= 54,Mx=Z50Zx20x2y dy dx =Z50h12x2y2iy=x2y=0dx =Z5012x4dx =110x5i50=542My=Z50Zx20x3dy dx =Z50x5dx =16x6i50=566,and so the center of mass is (256,12).5. The statement is true. On R the values of the function are bounded between −1 and1, and the area of R is 2. Therefore −2 = −1 × area(R) ≤RRRx sin(exy2+ x5) dA ≤1 × area(R) = 2.6. Using Fubini’s theorem,Z0−1Z1−x(e−y2+ xy) dy dx =Z10Z0−y(e−y2+ xy) dx dy =Z10hxe−y2+12x2yix=0x=−ydy=Z10ye−y2−12y3dy = −12e−y2−18y4i10= −12e+38.7. The volume is given byZ21Z40(4y − y2) dy dx =Z21h2y2−13y3iy=4y=0dx =323.38. The moment of inertia about the y-axis is given byIy=ZZZE(x2+ z2)xyz dV =Z10Z1−x0Zy2+2y2(x3yz + xyz3) dz dy dx=Z10Z1−x0h12x3yz2+14xyz4iz=y2+2z=y2dy dx=Z10Z1−x012x3y(y2+ 2)2+14xy(y2+ 2)4−12x3y5−14xy9dy dx=Z10h112x3(y2+ 2)3+140x(y2+ 2)5−112x3y6−140xy10iy=1−xy=0dx=Z10112x3((1 − x)2+ 2)3+140x((1 − x)2+ 2)5−112x3(1 − x)6−140x(1 − x)10−23x3−45xdx.Now expand to evaluate. The moment of inertia Ixabout the x-axis can be similarlycomputed asR10R1−x0Ry2+2y2(y2+ z2)xyz dz dy
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