1+(dy/dx) 2 dx. We get dy dx = 3x1/2 = 3 x, = 9x. Hence 1 +

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1 Mth.9 Em Solutions NAME: #.) / #.) / #.) /5 #.) / #5.) / #6.) /5 #7.) / Totl: / Instructions: There re 5 pges nd totl of points on the em. You must show ll necessr work to get credit. You m not use our book, notes, or clcultor. Prtil credit will onl be given for progress towrd correct solution. Once ou get our nswer down to numbers onl, ou do not hve to simplif..) points) Find the re of the region in the first qudrnt enclosed b the curves =, =, nd the -is. The first qudrnt is the region where > nd >.) =,) = The intersection of the curves occurs when =, or + =, or + ) ) =, hence = or = +. Since we re onl interested in the first qudrnt, = +, nd the intersection point is,). To find the re, it is esier to slice the region verticll, since ech verticl slice runs from = to =, nd hence hs length l) =. We hve one of these slices for ever between = nd =. So the re is A = l)d = d = ) = ) = 9 6 = 6..) points) Find the length of the curve = / from = to =. B the rc length formul, the length of the curve is +d/d) d. We get d d = / =, ) ) = 9. Hence + d d = + 9. Therefore the length of the curve is l = +9d. We so d d evlute this integrl using the substitution u = +9. Then du = 9d. When =, we hve u =, nd when =, we hve u = 7. Hence l = 7 9 u/ du = 9 7 u/ = 7 / /) = )

2 .) 5 points) The region in the first qudrnt enclosed b the curves = nd = is revolved round the -is. Find the volume of the resulting solid.,8) = = To find the intersection of the line = nd the curve =, we set = to get =. We solve this b fctoring: = or ) = or )+) =, so either =, =, or = +. Since we re looking onl t the first qudrnt, we hve = or = ; substituting into either eqution gives = when = nd = 8 when =, so the intersection points re,) nd,8). Since we re revolving the region round the -is, we slice this region verticll. We hve one of these slices for ever vlue of from = to =. For fied between = nd =, the slice revolved round the -is ields wsher with re πr r ). Here the outer rdius R of the wsher is the -vlue on the grph =, so R =. The inner rdius r is the -vlue on the grph =, so r =. Hence the re of the slice t is A) = π) ) ) = π6 6 ). Therefore the volume is V = A)d = = π 8 8 ) 7 π6 6 )d = π 6 ) = π = 8π 7) = 8π 7 = 8π 8 ) 7 7 = 5 π.

3 .) points) A wter tnk is is in the shpe of cone with the verte point) on the ground. The tnk hs rdius 5 ft nd height ft. The tnk is full of wter which weighs 6.5 lbs/ft. Find the work done, in ft-lbs, in pumping ll the wter out through point t the top of the tnk. Recll tht once our nswer onl hs numbers in it, ou cn stop. Constnts like π nd e count s numbers in this regrd. 5 R + We strt b setting up coordinte line with origin t the center of the top of the cone, with positive -is going down. To find the totl work, we will look t the wter in slice of width d t depth nd compute the work done in pumping tht slice up to the opening. To get the totl work done, we will dd up, vi n integrl, the work for these slices for ever strting t = nd ending t =. At prticulr depth, the work done in lifting the slice is the weight times the distnce. The distnce the wter t height must go is d =, b our choice of coordintes. The weight of the slice of wter t depth is the volume V) times the densit ρ = 6.5, so the weight is ρv). The volume of the slice with thickness d is V) = A)d, where A) is the re of the cross-section t depth. Since the slice is disk, its re is πr, where R is the rdius of the disk. To find R, we use similr tringles to s R tht = 5 =, so we get R = ) = 5. Putting this bck together, the re A) is πr = π 5 ). So the volume t height is V) = A)d = π 5 ) d. Hence the weight t height is ρv) = ρπ 5 ) d. Multipling b the distnce tht the wter t depth is rised, we get tht the work in rising the wter t depth to the top is ρπ 5 ) d = ρπ ) d = ρπ ) d. Hence the totl work done in pumping ll the wter out is W = ρπ ) 5 d = ρπ 5 + ) 6 =,5ρπ + ) 6 8+ =,5ρπ =, 56.5)π ),5 = ρπ 5, ),96 ft-lbs. +, ) 6

4 5.) points) Find the verge vlue of the function f) = sin on the intervl [π,π]. b B the formul f)d for the verge vlue of function f on n intervl [,b], in this cse b the verge vlue is A = π Hence π π sin d. We evlute this integrl using the identit sin = cos). π A = π π cos)d = π ) π sin) π π) sinπ) π )) sinπ) = π since sinπ) = nd sinπ) = = π π π ) = π π =, 6.) 5 points) Find the centroid,) of the region in the first qudrnt bounded bove b the curve =, below b the -is, nd on the right b =. We use the formuls = f)d f)d nd = f)) d f)d, where f) =, =, nd b =. =.,) = nd We clculte f)d = f)d = ) d = / d = / = / ) = d = / d = ) f)) d = d = d = 8 = 6, ) 5 5/ = 5 5/ ) = 5 = 6 5, ) = 6 Hence = 6/5 6/ = = 5 nd = 6/ = 6 =. Hence the centroid of the region is,) = 5, ) =.,.75). =.

5 7.) points) The region enclosed b the curves =, = nd the -is is revolved round the line =. Find the volume of the resulting solid.,) = = = For the curve =, we hve = when =. Since we re revolving round the line =, we slice this region horizontll. We hve one of these slices for ever vlue of from = to =. For fied between = nd =, the slice revolved round the line = ields disk with re πr. Here the rdius R of the disk is the -vlue on the grph =, minus the -vlue on the grph =, or =. Tht is, R =. Hence the re of the slice t height is A) = π ) = π +) = π / +). Therefore the volume of the solid is V = A)d = π / + ) d = π ) / + = π 6 8 ) 8+8 = π 6 ) = π 7 6 = 8 π. 5