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MATH2400: CALCULUS 3
MAY9,2007
FINAL EXAM
I have neither given nor received aid on this exam.
Name:
001 E.Kim ................(9am)
004 M.Daniel ........... (12am)
002 E.Angel .............(10am)
005 A.Gorokhovsky ......(1pm)
003 I. Mishev ............ (11am)
If you have a question raise your hand and remain seated. In order to receive full credit your answer
must be complete, legible and correct. Show all of your work, and give adequate explanations.
DONOTWRITEINTHISBOX!
Problem Points Score
1 15 pts
2 15 pts
3 15 pts
4 15 pts
5 20 pts
6 15 pts
7 30 pts
8 15 pts
9 15 pts
10 30 pts
11 15 pts
TOTAL 200pts
1. (15 pt) Find the area of the region enclosed by the curve
1. (15 pt) Find the area of the region enclosed by the curve
2 2
x = 1−t2, y = t(1−t2), −1 ≤ t ≤ 1
x=1−t , y =t(1−t ), −1≤t≤1
(see the picture below)
(see the picture below)
0.5
0 0.5 1
-0.5
ByGreen’s Theorem, A = I −ydx with C the above curve oriented counter-clockwise. Thus
I Z C Z
1 1 8
2 2 4
A= C−ydx= −1−t(1−t )(−2t)dt=2 −1(t −t )dt= 15.
2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2x2 + 3y2 that is also
parallel to the plane 4x − 3y − z = 10
2. (15 pt) Find the equation for the plane tangent to the paraboloid z = 2x2 + 3y2 that is also
parallel to the plane 4x − 3y − z = 10
Write G(x,y,z) = 2x2+3y2−z, so ∇G(x,y,z) = h4x,6y,−1i. This tells us that h4x,6y,−1i
is a normal vector for the tangent plane to our surface at the point (x,y,z). For the tangent
plane at a point to be parallel to 4x − 3y − z = 10, we need h4x,6y,−1i to be parallel to
h4,−3,−1i. This will certainly happen when x = 1 and y = −1. Solving for z in the equation
1 11 2
for our surface, we get that the tangent plane at (1,−2, 4 ) is parallel to 4x−3y−z = 10. At
this point the equation for the tangent plane will be given by 4(x−1)−3(y+1)−(z−11) = 0.
2 4
3. (15 pt) Find the flux of F(x,y,z) = (x + y)i + (y + z)j + (z + x)k across the portion of the
plane x+y+z =1 in the first octant oriented by unit normals with positive components.
Weknow Φ=RR F•ndS. Let u=x and v =y. Then our surface is given by
σ
r(u,v) = hu,v,1−u−vi, 0 ≤ u ≤ 1, 0 ≤ v ≤ 1−u.
Next,
ndS =(r ×r )dA=(h1,0,−1i×h0,1,−1i)dA=h1,1,1idA.
u v
Then
Φ=ZZ F•ndS
σ
=Z 1Z 1−uhu+v,v+(1−u−v),(1−u−v)+ui•h1,1,1idvdu
0 0
=Z 1Z 1−u2dvdu
0 0
=1
4. (15 pt) Evaluate the integral Z Z
1 1
cos(y3)dydx
0 √x
by changing the order of integration.
Z 1Z 1 cos(y3)dydx = Z 1Z y2 cos(y3)dxdy
0 √x 0 0
=Z 1cos(y3)y2dy
0
1 1
3
= sin(y )
3 0
=sin(1)
3
5. A wire 12cm long is cut into three or fewer pieces, with each piece bent into a square.
(a) (10 pt) What is the minimal total area of the squares? Justify your answer.
Let x,y,z denote the lengths of the three segments of the wire after it is cut, so x+y+z =
12. Bending each segment into a square yields a total area of A = x2 + y2 + z2.
4 4 4
Replacing z with 12−x−y (and simplifying), we have
A(x,y) = 1(x2 +y2 +xy −12x−12y+72).
8
Now, x+y+z =12forces x and y to be confined to the region in the xy-plane, R, which
is the triangle with vertices (0,0),(12,0),(0,12) (this is just the projection of surface
x+y+z=12,withx,y,z ≥0 onto the xy-plane). We now minimize and maximize A
over R.
Find Interior Critical Points: We simultaneously solve A = 1(2x+y−12) = 0 and
x 8
A = 1(2y+x−12)=0tofindthe one interior critical point (4,4,4).
y 8
Test Critical Points and Boundary Points:
On the portion of the boundary of R that lies on the x-axis, we have y = 0 and
0 ≤ x ≤ 12. Substituting into A we get A = 1(x2 − 12x + 72), 0 ≤ x ≤ 12. The
1 8
absolute extrema of (x2 − 12x + 72) on 0 ≤ x ≤ 12 occur when x = 0,12, so we must
8
consider the points (0,0,12),(12,0,0).
OntheportionoftheboundaryofRthatliesonthey-axis,wehavex = 0and0 ≤ y ≤ 12.
Similar to the previous case, we find that we must consider the points (0,0,12),(0,12,0).
Finally, on the portion of the boundary of R that lies on the line y = 12 − x, we have
y = 12−xand0≤x≤12. SubstitutingintoAwegetA= 1(x2−12x+72),0≤x≤12.
1 8
As before, the absolute extrema of (x2−12x+72) on 0 ≤ x ≤ 12 occur when x = 0,12,
8
so we must consider the points (0,12,0),(12,0,0).
2
Testing all the points that we have listed, we find that the minimum total area is 3cm
occurring when each segment is of length 4cm.
(b) (10 pt) What is the maximal total area of the squares? Justify your answer.
2
Using the above information, we find that the maximum total area is 9cm occurring
when we make no cut and have only one segment of length 12cm.
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