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PRACTICE PROBLEMS: SET 4 MATH 101: PROF. DRAGOS GHIOCA 1. Problems Problem 1. Compute the area of the region bounded by y = cos(x), y = cos(2x), x = π2 and x = π. Problem 2. Find the volume of the solid obtained by rotating the region between the curves y = x4 and x = y 4 around the line y = 5. Problem 3. Find a positive number c such that the region between the curve y = c − x2 and the x-axis has area equal to 5. 1 2 MATH 101: PROF. DRAGOS GHIOCA 2. Solutions Problem 1. The area of the region between the graphs of cos(x) and of cos(2x) bounded by x = π2 and x = π is ∫ π |cos(x) − cos(2x)| dx . π 2 So, we need to determine which function (cos(x) or cos(2x)) is larger on each subinterval of [π/2, π]. So, for this we solve what are the points of intersection for the two graphs. Therefore, we equate: cos(x) = cos(2x). In order to solve the above equation we note the identity: cos(2x) = 2 cos2 (x) − 1, and thus we have to solve: cos(x) = 2 cos2 (x) − 1, which yields 2 cos2 (x) − cos(x) − 1 = 0, and hence (2 cos(x) + 1)(cos(x) − 1) = 0. [ ] So, we obtain either cos(x) = − 12 or cos(x) = 1. We note that on the interval π2 , π the cosine only takes values less than or equal to 0, and thus we only have to solve 1 cos(x) = − . 2 [ ] The unique solution x ∈ π2 , π is x = 2π . More precisely, 3 ( ) 1 π 2π cos(x) > − for x ∈ , 2 2 3 and therefore (since always cos(x) − 1 < 0), we get ( (2 cos(x) + 1)(cos(x) − 1) < 0 for x ∈ Similarly, π 2π , 2 3 ( (2 cos(x) + 1)(cos(x) − 1) > 0 for x ∈ which in particular yields that ( cos(x) < cos(2x) for x ∈ and cos(x) > cos(2x) for x ∈ ∫ π |cos(x) − cos(2x)| dx π 2 (π 2π 2, 3 ∫ 2π 3 = π 2 2π ,π 3 ) . ) 2π ,π , 3 ) ) . Therefore the desired area equals ∫ cos(x) − cos(2x) dx + π 2π 3 cos(2x) − cos(x) dx . In order to evaluate (using the Fundamental Theorem of Calculus) the above definite integrals we need to find the antiderivative of cos(2x). We do this using the substitution u = 2x and so, du = 2 dx . PRACTICE PROBLEMS: SET 4 Thus ∫ ∫ 3 cos(u) sin(u) sin(2x) du = +C = +C. 2 2 2 cos(2x) dx = We finally compute ∫ π |cos(x) − cos(2x)| dx π 2 ( ( ) ) 2π sin(2x) sin(2x) sin(x) − − sin(x) |π2π | π3 + 2 3 2 2 ( √ (√ √ ) √ ) 3 3 3 3 + − (1 − 0) + (0 − 0) − − − 2 4 4 2 √ 3 3 −1 2 √ 3 3−2 . 2 = = = = Problem 2. We first find the points where those two curves intersect. So, we have x = y 4 = (x4 )4 = x16 . From the equation x = x16 we obtain that either x = 0 or x15 = 1 in which case we have x = 1. So, the two curves intersect in two points: (0, 0) and (1, 1). √ We note that for x ∈ [0, 1], the curve x√= y 4 is the same with the curve y = 4 x. Moreover, on this interval, we have that 4 x ≥ x4 . When we rotate the region around the line y = 5, we obtain the general sections, when we slice the region at a point x between 0 and 1, are annuli (i.e., region between two circles). The radius of the large circle is the distance between the curve y = x4 and the line y = 5, and thus, it is 5 − x4 . On the √ other hand, the 4 radius of the small circle is the distance between the curve y = x and the line √ y = 5, and so, it is 5 − 4 x. In conclusion, the volume of the solid is ∫ 1 ( √ ) π · (5 − x4 )2 − (5 − 4 x)2 dx 0 ∫ 1 = 0 ( √ √ ) π · 25 − 10x4 + x8 − 25 + 10 4 x − x dx ∫ ( ) 1 1 π · x8 − 10x4 − x 2 + 10x 4 dx 0 ) ( 3 5 x9 x2 x4 5 =π· − 2x − 3 + 10 · 5 |10 9 2 4 ) ( 2 1 −2− +8 =π· 9 3 49π = . 9 1 = Problem 3. The parabola − x2 is symmetric with respect to y-axis. More√y = c √ over, its x-intercepts are − c and c (note that it is given to us that c is positive). 4 MATH 101: PROF. DRAGOS GHIOCA So, the area of the region is double the area enclosed by the x-axis, y-axis and the curve y = c − x2 . In other words, the area equals ∫ √c 2· c − x2 dx 0 ( ) x3 √c = 2 · cx − |0 3 ( √ ) √ ( c)3 =2· c· c− 3 3 = 4c 2 . 3 So, we need to find c such that 3 4c 2 = 5, 3 3 which yields c 2 = 15 4 and so, ( c= 15 4 ) 23 .