Review (2) Calculus II (201-nyb-05/05,06) Winter 2019

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1 Review () Calculus II (-nyb-5/5,6) Winter 9 Note. You should also review the integrals eercises on the first review sheet. Eercise. Evaluate each of the following integrals. a. sin 3 ( )cos ( csc 3 (log)cot 5 (log) ) d b. d c. sin 6 (ϑ) dϑ d. (5 z z ) e. + d f. (tan ) d 3 (y 5) 3/ g. y 5 h. + 3 d i. 3u + 5u + du j ( d k ) + d l. 3 y + y + 3 m. (y + ) y + 3 n. p. y /3 ( + y /3 ) 3/ q. 6 d s d d o. ( )( + 5) d sec 3 (ϑ)tan sec 6 ( w)tan ( w) (ϑ) dϑ t. dw u. w v. e y + 6e y 7 r. log( + ) d 6 dϑ sinϑ w. 3 ( ) ( ) 7 d. e cos() d Eercise. Evaluate each integral, or show that it diverges. Be specific. If an integral diverges to or to, then give this information as well, together with a brief justification. a. d. (log) d b. cos log(sin) sin dw g. w 6 w j. arctan 9 p p dp c. p 3y + y y 8 arcsecz arccsc d e. z 3 f. 3 d h. + d i. e cos() d d k. + 3 d l. ( )( + )( ) Eercise 3. Determine all values of r for which the improper integral sin r d is convergent. 3 ( ) d 3 Eercise. Compute the area of the region enclosed by the graphs of y = ( )y =. Eercise 5. Epress the area of the region enclosed by the graphs of y = 5( ) ( ) y = (+)( 5+6)( +3 ) as the sum of definite integrals of polynomials. Use as few integrals as possible. Eercise 6. The region R enclosed by the graphs of y = + y = 3 3 is sketched below. y y = + y = 3 3 Let R denote the part of R to the left of the y-ais, let R denote the part of R so the right of the y-ais. Write an integral which is equal to a. the area of R, b. the volume of the solid obtained by revolving R about the line defined by y =, c. the volume of the solid obtained by revolving R about the line defined by = d. the length of the perimeter of R. Evaluate the integrals in parts a c. Eercise 7. The functions f g are continuous on [,7], where f () g(), F denotes the figure enclosed by their graphs the lines defined by = = 7. If F is revolved about the line defined by y =, the volume of the resulting solid is 69. If F is revolved about the line defined by y = 5, the volume of the resulting solid is 39. Determine the area of R. Eercise 8. Find the length of each curve. a. = log( y ), from (,) to ( log 3, ). b. y = log, for. c. y = log(cos), for 3. Eercise 9. Let R be the region enclosed by the graphs of y = cos, y =, = =. Compute the area of R, find the volume of the solid obtained by revolving R about the -ais, the y-ais the line defined by y =. Eercise. The boundary inside of a four dimensional sphere of radius r is the set of all points (,y,z,t) such that + y + z + t r. Using the idea of integrating cross sections, compute the volume of a four dimensional sphere of radius r. Eercise. Give the eplicit solution to each initial value problem. a. d sec() = y cos(); y( ) = e. b. e t ( t) d dt = t( ) ; () =. Eercise. Show that the solutions of the differential equation d y d = d + 6y are precisely those functions of the form Ae + Be 3, where A B are real numbers.

2 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 Solution to Eercise. a. If t = cos( ), then dt = sin( ) d. Also, sin ( ) = t, t = if = t = if =. Hence (using symmetry in the second equation) sin 3 ( )cos ( ) d = ( t )t dt = (t t 6 ) dt = ( 5 t 5 7 t7) = 35. b. If t = csc(log), then dt = csc(log)cot(log) d, so csc 3 (log)cot 5 (log) d = t (t ) dt = (t 6 t + t )dt = 7 csc7 (log) + 5 csc5 (log) 3 csc3 (log) + C. c. Applying the half-angle identity for the sine function gives ( sin 6 (ϑ) dϑ = 8 ( cos(ϑ)) 3 dϑ = 8 ) 3cos(ϑ) + 3cos (ϑ) cos 3 (ϑ) dϑ = 8 ϑ 3 3 sin(ϑ) cos (ϑ) dϑ 8 cos 3 (ϑ) dϑ. Net, applying the half-angle identity for the cosine function to the first remaining integral gives cos (ϑ) dϑ = ( + cos(8ϑ)) dϑ = ϑ 6 sin(8ϑ) + C, changing the variable of integration to t = sinϑ in the second remaining integral yields cos 3 (ϑ) dϑ = ( t ) dt = t t3 + C = sin(ϑ) sin3 (ϑ) + C. Combining the foregoing results gives sin 6 (ϑ) dϑ = 6 5 ϑ 8 sin(ϑ) + 96 sin3 (ϑ) 8 3 sin(8ϑ) + C. Note: Alternatively, one could use the reduction formula for powers of sine: sin 6 (ϑ) dϑ = 6 sin 5 (ϑ)cos(ϑ) 5 6 sin3 (ϑ)cos(ϑ) sin(ϑ)cos(ϑ) (ϑ) + C = sin5 (ϑ)cos(ϑ) 5 8 sin3 (ϑ)cos(ϑ) 5 3 sin(ϑ)cos(ϑ) 5 6 ϑ + C. d. If y = z 5 z, then y = 5z, so 5 y = z 3 z = 5 (y ). Hence, ( ) z (5 z ) 3 = z 3 5 z z z 3 = y 3 y y 5 5 = ( + y ) = ( y y ) + C = y 5 5 5y + C = ( 5z )z 5 5 z + C = z 5 5z 5 z + C. e. If t = + then = (t ) d = t(t )dt; therefore, + d = + d = t (t ) 3 dt = (t 8 3t 6 + 3t t ) dt = 9 t t t5 3 t3 + C = 35 t3 (35t 6 35t 89t 5) + C. To epress the result in terms of in simplified form, observe that so t 6 = , t = + + t = +, 35t 6 35t 89t 5 = d = 35 ( ) ( + ) 3 + C. f. As the notation tan is repulsive, arctan will be written instead. Partial integration, taking the primitive ( ) of, gives (arctan) d = ( )(arctan) = ( )(arctan) ( ) arctan d arctan d. Now integrating by parts by inspection gives arctan d = arctan d = arctan log( ) + C. (arctan) d = ( )(arctan) arctan log( ) + C. g. If y = 5secϑ, then = 5 secϑ tanϑ dϑ, y 5 = 5tanϑ, hence (y 5) 3/ (5tanϑ) 3 ( 5 secϑ tanϑ ) y 5 = ( 5 secϑ ) dϑ = sin (ϑ) dϑ. The half angle identities for the sine cosine functions give sin (ϑ) = ( ) cos(ϑ) + cos (ϑ) = 3 8 cos(ϑ) 8 cos(ϑ). 6 5 sin (ϑ) dϑ = 5 6 ϑ 5 sin(ϑ) sin(ϑ) + C. The double angle identities for the sine cosine function imply that 5 sin(ϑ) sin(ϑ) = 5 (cos(ϑ) )sin(ϑ) = 5 ( ) cos (ϑ) 5 sinϑ cosϑ. Now ϑ = arcsec ( 5 y ) y 5, sinϑ =, cosϑ = 5 ( ) 5 y y cos ϑ 5 = 5 = 5(5 y ) y y ; hence, (y 5) 3/ y 5 = 6 5 arcsec( 5 y ) 5(5 y ) y 5 + y + C.

3 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 Note. The integral in question could be evaluated by repeated partial integration. Integrating the power of y differentiating the radical epression at each stage gives (y 5) 3/ y 5 = (y 5) 3/ y 5 y + 3 y 3 = (y 5) 3/ y 3 y 5 y + 6 y y 5 5(y 5) y 5 = y arcsec( 5 y ) + C. h. If y = + 3 = ( + ) + 9 then ( + ) = y 9 ( + ) d = y, so + 3 y ( ) d = + y 9 = 9 + y = y log y 3 + C y + 3 = log C. i. Partial integration, using the primitive 6 (6u 5) of u, noting (3u 5u+) = (6u 5) +3, gives 3u 5u + du = 6 3u (6u 5) 5u + (6u 5) u du 5u + = 6 3u (6u 5) 5u + 3u 5u + du du (6u 5) + 3 = 3u (6u 5) 5u ( 3log 6u u 5u + ) + C, where the second term on the second line is absorbed by the left side. j. If t = + then ( + + 7) d = t (t + 3) dt = dt t (t 3 + 3) = 6 + dt (t + 3). Adding subtracting t in the numerator then integrating by parts gives dt (t + 3) = dt t + 3 t t (t + 3) dt = dt t t 3 3 dt (t + 3) (t + 3) = + 6 ( 3arctan 3 t 3 ) 3 = 3. the integral in question is equal to 6 3 ( 3 ) = k. Since + = ( )( + ), there are rational numbers a, b, c d such that Clearing denominators gives a + b + c + d + = (a + b)( + ) + (c + d)( ) = , comparing the cubic, quadratic, linear constant coefficients gives, respectively, a + c = 3, a + b c + d =, a + b + c d = 6 b + d =. Subtracting the first equation from the second third equations gives b c + d = 3 b d = 3. Subtracting the fourth equation from the fifth sith equations gives c =, or c =, d =, or d =. The first fourth equations then give a = b =. Now = + ( ) = ( ) + 3, so d = d + 8 d ( ) + 3 = log( ) + ( ) 3 3arctan + C. 3 Also, = ( ) 5 ( + ) = ( ) + 3, so + d = + d d ( ) + 3 = log( + ) 3 5 ( ) 3arctan + C. 3 the integral in question is equal to log ( ( ) + ) + ( ) 3 3arctan 5 ( ) 3 3 3arctan + C. 3 l. Division gives = , + is a primitive of the quotient. Net, is a zero of multiplicity two of the denominator, factorizing by inspection gives = ( ) ( +). The resolution into partial fractions of the proper part has the form + ( ) ( ) = a ( ) + b + c, in which the coefficient over ( ) is obtained by inspection (covering evaluating). Clearing denominators gives + = a( )( ) + ( ) + (b + c)( ), comparing the cubic, quadratic linear coefficients gives, respectively, a + b =, a b + c = a + b c =. Using the first equation to eliminate a from the second third equations yields b + c = c =, so c =, b = a =. the integral of the proper part is ( ) + d = log + log( ) + arctan + C d = + + log + arctan + C. ( ) m. If t = y + 3 then y = (t 3), = t dt y + = (t ). Hence, y + y + 3 (y + ) y + 3 = (t 3) + t t + t 3 t dt = (t )t t dt = + t t dt = t + log(t ) arctan(t) + C = y log(y + ) arctan( y + 3) + C.

4 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 n. If y = then 6 = y 9 y = 6 d. So d(6 3 y) = 8 y d 6 3 = 8 y d + y 9 = 6 y d 9, d(y) = y d + = 9 y = 9 d +. y d + 6 y d + the integral in question is equal to ( ) y d = 3 y = 3 y 8 9 d y 9 y = 8 y(3 + 9) 5 8 log( + y) + C = 8 (3 + 9) log( ) + C. o. The resolution into partial fractions of the integr has the form ( )( + 5) = a + b + c D + E ( + 5), in which a = by inspection (covering evaluating). Clearing denominators yields = a( + 5) + (b + c)( )( + 5) + (d + e)( ) then comparing the coefficients of powers of in decreasing order gives: a + b =, so b =, a b + c = 9, so c =, a + 3b c + d = 3, so D = A + 5B + 3C + D + E = 9, so E = 3. the integral is equal to ( + 5) d. A primitive of the first term is log( ). Net, + 5 d = + 5 d + 3 d + 5 = log( + ) + 3 arctan( ( ) ) + C, 3 ( + 5) d = ( + 5) d d ( + 5) = ( + 5) Adding subtracting t in the numerator integrating by parts gives dt dt (t + ) = t + t t (t + ) dt = t (t + ) dt t +. Thus, with t = this last calculation shows that d ( + 5) = ( + 5) 8 arctan( ( ) ) + C. Combining these results gives, ( )( + 5) d = ( + 5) + log( ( ) ( + 5) ) 8 arctan( ( ) ) + C. d ( + 5). p. If t = y /3 ( + y /3 ), then t = y /3, 3t dt = y 5/3 t 3 (t ) = y /3 ( + y /3 ) 3/. y /3 ( + y /3 ) 3/ = 3 t (t ) dt = 3 ( t ) dt = 3(t + t ) + C = 3(t ) t = 3( + y/3 ) y /3 ( + y /3 ) + C. + C = 3(y /3 + ) y /3 ( + y /3 ) + C q. If t = 3, then 6 = 9 t, = (t + 3) = t + 6t + 9. Partial integration gives t 9 9 t dt = t t + 9 t dt = t 9 t dt t t 9 t dt = 9 t t + 9 dt 9 t, by inspection, 6t 9 t dt = 9 t + C. 6 d = (t ) 9 t + 7 dt 9 t = ( + 9) arcsin( 3 ( 3) ) + C. Note. It is also possible to use the change of variables 3 = 3sinϑ. In this case, d = 3cosϑ dϑ 6 = 3cosϑ. Hence, Now so ( 6 d = 3( + sinϑ) ) 3cosϑ cosϑ = 3 = 9ϑ 8cosϑ + 9 3cosϑ dϑ = 9 ( + sinϑ + sin ϑ)dϑ ( cosϑ) dϑ = 7 ϑ 8cosϑ 9 sinϑ + C = 7 ϑ 8cosϑ 9 sinϑ cosϑ + C. 6, sinϑ cosϑ = 9 ( 3) 6, d = 7 6 arcsin( 3 ( 3) ) 6 6 ( 3) 6 + C r. Integrating by parts gives = 7 arcsin( 3 ( 3) ) ( + 9) 6 + C. log( + ) d = ( )log( + ) ( ) + d = ( )log( + ) 7 ( ) + 7 d = ( )log( + ) + 7arctan ( 7 ( ) 7 ) + C.

5 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 s. Since d sec 3 (ϑ)tan(ϑ) = 3sec 3 (ϑ)tan (ϑ) + sec 5 (ϑ) = sec 3 (ϑ)tan (ϑ) + sec 3 (ϑ), d d sec(ϑ)tan(ϑ) = sec(ϑ)tan (ϑ) + sec 3 (ϑ) = sec(ϑ) + sec 3 (ϑ), d it follows that d sec 3 (ϑ)tan(ϑ) sec(ϑ)tan(ϑ) = 8sec 3 (ϑ)tan (ϑ) + sec(ϑ). d Rearranging this last equation gives sec 3 (ϑ)tan (ϑ) dϑ = sec3 (ϑ)tan(ϑ) 8 sec(ϑ)tan(ϑ) 8 log sec(ϑ) + tan(ϑ) + C. t. If t = tan( w), then dt = w / sec ( w) dw sec ( w) = (t ) = t + t. Hence, sec 6 ( w)tan ( w) dw = t (t + t ) dt = (t 6 + t + t ) dt w u. If z = tan ( ϑ ), then sinϑ = If ϑ = then z =, if ϑ = 6 then = 7 tan7 ( w) + 5 tan5 ( w) + 3 tan3 ( w) + C. z + z = + z. z = tan ( ) = sin( 6 ) + cos ( 6 ) = + 3 = 3. The inverse tangent is an odd function, so the last calculation implies that arctan( 3 ) =. 6 dϑ sinϑ = z = z +z = ( 3 3 arctan( 3 ) arctan ) 3 3 = 3 3 ( 6 ) = 8 3. (z ) + 3 = 3 ( ) z 3 3arctan 3 v. If z = e y then logz = y z =, so e y + 6e y 7 = z 3 7z + 6. The denominator of the integr vanishes if z =, so it is divisible by z, factorizing by inspection gives z 3 7z + 6 = (z )(z + z 6) = (z )(z )(z + 3). Resolving the integr into partial fractions by inspection (i.e., covering evaluating) then integrating gives therefore, z 3 7z + 6 = (z ) 5(z ) (z + 3) = log (z ) (z + 3) (z ) 5 e y + 6e y 7 = log (e y ) (e y + 3) (e y ) 5 + C. + C; w. By elementary properties of eponents, 3 ( ) ( ) 7 d = Let t be the radical factor of the last integr, so that 3 ( )( ) d. t 3 =, t d dt = ( ), = t3 t 3, = 3t3 t 3 = t6 + t 3 (t 3 ). t 3 3 ( ) ( ) 7 d = 3t 3 t6 + t 3 (t 3 ) t ( t )dt = t 6 + t 3 3 t 3 dt. Division resolution into partial fractions gives t 6 + t 3 t 3 = t Now t + t + t dt = t t + t dt (t )(t + t ) = t (t + ) t t + t. () dt (t ) + 3 = log(t + t ) + ( ) t 3arctan + C. 3 Integrating the remaining terms of () by inspection then combining the results gives 3 ( ) ( ) 7 d = 3 (t + 8t) log t + t (t ) + ( ) t 3arctan + C. 3 The polynomial function of t is 3 (t )t = 3( ) = (3 5) 3 ( ) /3. The argument of the logarithm is the absolute value of t 3 (t ) 3 = 3 ( ) = ( 3 3 ) 3, so the logarithmic term of the integral is 3 log 3 3. The numerator of the argument of the inverse tangent is 3 ( ) ( ) 7 d. Since it follows that d d t = 3 = = (3 5) 3 ( ( ) /3 3 log ) + 3arctan + C. 3 3 d e cos() = e cos() + e cos() e sin() d d e sin() = e sin() + e cos() + e sin(), d e (cos() + sin()) = e cos() + e (cos() + sin()) = e cos() + d d e sin().

6 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 e cos() d = e( (cos() + sin()) sin() ) + C. Solution to Eercise. a. Partial integration gives (log) d = (log) log d = (log) 8(log) d + 8 = (log) 8(log) 6 + C. Since ε p (logε) n as ε + if p > n is a positive integer, it follows that (log) d = lim ε + ε (log) d = 6 lim ε + (logε) ε 8(logε) ε 6 ε = 6. b. Changing the variable of integration to ϑ = arcsec( p), so that p = sec ϑ, gives dp = sec ϑ tanϑ dϑ, Also, ϑ + as p + ϑ as p. p p p dp = p = secϑ p = tanϑ. (secϑ )(sec ϑ tanϑ) sec dϑ = (cosϑ cos ϑ) dϑ. ϑ tanϑ The discontinuities of the integr are removable, so the resulting integral is (essentially) definite. p p p dp = cosϑ dϑ cos ϑ dϑ = (sinϑ) ( + cosϑ) dϑ = ϑ sinϑ =. c. If < y 5, then 3y + > y + 7, so 3y + y y 8 = 3y + (y + )(y ) > y > y >, 5 y = is divergent (p = in the scale of powers at the origin), so the improper integral is divergent, by the comparison principle. 9 3y + y y 8 d d. Since d d log(sin) = cos sin, integrating by inspection gives cos log(sin) sin d = lim ε + ε cos log(sin) sin since sinε +, therefore log(sinε), as ε +. e. Partial integration gives t arcsecz arcsecz z 3 = lim t z + d = lim ( ) log(sin) ε + =, ε z 3 z = z 3 z, since < (arcsect)/(t ) < t if t >, t as t. If y = z then z = y, y + as z +, y as z, y = z ( ) y d z = z y = z3 z (z ) z 3 = y z 3 y y. the integral in question is equal to z 3 y = lim y y (y ) y = y lim arctan(y) = 8. f. If ϕ = arccsc, then < ϕ, so < = sinϕ < ϕ = arccsc, hence arccsc 3 > 3 = >. 5/6 But d 5/6 is a divergent improper integral (p = 6 5 < in the scale of powers at ), so the improper integral arccsc 3 d is divergent, by the comparison principle. g. If = w w, then = w d = w 3 dw. If w + then + if w then, so the impropriety is resolved by the change of variables. Since w = ( ), the integral in question is equal to w w w dw w 3 = 8 ( ) d = ( ) = ( ) = 5. h. Since α d = α lim d + lim d β + = lim arcsecα lim α arcsecβ = β +, β if t =, then γ d = γ lim dt t = γ lim log t γ = t log = log( + ),

7 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 it follows that + d = d + d = + log( + ). i. Partial integration (integrating the eponential factor) gives e cos() d = e cos() + e sin() e cos() d = e (sin() cos()) + C. α e cos() d = lim α e cos() d = lim α since e α (sinα cosα) e α, e α as α. j. The integral is a sum of limits of definite integrals as follows. arctan() d = lim δ + δ arctan() ω d + lim ω Partial integration (integrating the power) gives arctan() d = arctan() + If ω > then sin(α) cos(α) e α =, arctan() d ( + ) < arctan(ω) < arctanω, thus lim = because lim ω ω ω ω ω Also, if δ > then < arctanδ <, so < arctanδ < tan(arctanδ) = δ, hence < arctanδ δ d. =. ω < δ = arctanδ δ, so lim =, because lim δ δ + δ δ +( δ) =. arctan d = d ( + ). If y = (), then = y d = y. Also y as + y as, so d ( + ) = y ( y ) y = 8 y +, where now the discontinuity at the origin is removable. The denominator of the integr factorizes as y + = (y + y + )(y y + ), in the resolution into partial fractions, ay + b y + y + + cy + d y y + = 8 y +, Subtracting the first equation from half the third equation gives b +d =, one-half of the fourth equation is b + d =, so b = d =. The second equation now gives a + c =, so a = c = using the first equation. Now y + y + y + = y y y + = arctan y y + y + + y y + y + (y ) = log(y + y + ) + arctan(y ) + C, (y ) = log(y y + ) arctan(y ) + C. d = 8 y + = lim log y + y + α α y y + + arctan(y ) + arctan(y ) = ( ) ( ) =. k. Resolving the integr into partial fractions, integrating, collecting the logarithms, gives + 3 ( )( + )( ) d = / /3 d = 3 lim log ( ) 7 β ( ) 6 ( + ) = 3 log 6 log 3 6 = log ( 9 3 ). β = 93 3 log l. The denominator of the integr is = ( ) ( + ), so the integr has infinite discontinuities at, each of which falls in the interval of integration (so the improper integral is the sum of four limits of definite integrals). If <, then 3 < 5 3 < +, so 3 ( ) = ( + ) /3 ( ) /3 > 3 /3, ( ) /3 d ( ) /3 = d /3 is a divergent improper integral (p = 3 > in the scale of powers at ), so the comparison principle implies that improper integral 3 d, therefore also ( ) 3 d, ( ) is divergent. (The indefinite integral could have been evaluated using a rationalizing change of variables, but in this case that is not necessary.) clearing denominators gives (ay + b)(y y + ) + (cy + d)(y + y + ) = 8. Comparing coefficients gives a + c =, a + b + c + d =, a b + c + d = b + d = 8.

8 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 Solution to Eercise 3. First notice that if r k is a positive integer then r sin sin on [k(k ) ], so direct evaluation of the corresponding definite integral of the sine function gives (k+) k sin r d if k is even, if k is odd. So if r, then the integral in question (oscillates ) diverges. On the other h, recall that so It follows that cos < sin <, if < <, < sin, therefore sin < r r, if < < 3. sin r d diverges with d r by the comparison principle (using the scale of powers near the origin) if r (since then r ). So the integral in question diverges if r or r. Now write Since it follows that sin sin r d = <, therefore sin r d + sin r d converges with sin r d. sin r < r, if < <, d r by the comparison principle (using the scale of powers near the origin) if r < (since then r < ). Net, observe that if r > then sin r d = lim t cos r t r cos r+ d = r cos d r+ by partial integration (as r cos r if >, r as ). Now (r+) cos (r+) if >, so cos d d converges with r+ r+ by the comparison principle (using the scale of powers at ) if r > (since then r > ). Therefore sin r d converges if r >. It follows that is convergent if, only if, < r <. sin r Solution to Eercise. The curves meet where = = + = ( )( + + ). Since + + = ( ( ) + 7 ), has no real zeros, the curves meet at the points (,) (,). If < < then, by the displayed factorization, <. the area of the region enclosed by the given curves is equal to Solution to Eercise 5. Let then d = log( ) 3 3 = log 3. y = ( )( 5 + 6)( + 3 ) y = 5( ) ( ); y y = ( )( 5 + 6)( + 3 ) 5( ) ( ) d = ( + 5)( )( ) ( 3) 5( )( )( ) = ( )( ) ( 3 ) = ( + )( )( ) ( 5). the curves meet where is,, 5. Analyzing the sign of the factors of y y reveals that y > y if < <, y y if < < 5. the area of the region enclosed by the curves is equal to or (y y ) d + ( + )( )( ) ( 5) d + Solution to Eercise 6. The curves meet where 5 (y y) d, 5 ( + )( )( ) (5 ) d. + = 3 3, or ( + )( ) =. If < <, the quadratic function is larger, if < <, the cubic function is larger. a. The area of R is equal to ( 3 + ) d + ( 3 + ) d = = = 37.

9 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 b. If R is revolved about the line defined by y =, then the cross sections perpendicular to the -ais of the generated solid are annuli of inner radius outer radius, for. So the volume of the solid is equal to ( ) ( ) d. c. If R is revolved about the line defined by = then the resulting solid consists of concentric cylindrical shells of radius height 3 +, for. So the volume of the solid is equal to ( )( 3 + ) d = ( + 3 ) d = = d. The length of the perimeter of R is equal to the sum of the lengths of the curves on [,], or + ( ) + + 9( ) d. Solution to Eercise 7. The volume of the solid obtained by revolving R about the line defined by y = 5 is 39 = 7 (5 f () ) ( 5 g() ) d = 7 ( )( ) g() f () f () g() d. The volume of the solid obtained by revolving R about the line defined by y = is 69 = 7 (g() ) ( f () ) d = 7 ( )( ) g() f () f () + g() + d. The sum of these volumes is 7 ( ) 7 ( ) 8 = g() f () d, so g() f () d = 9 is the area of R. Solution to Eercise 8. d = y y, so ( ds a. If = ln( y ), then the length of the curve in question is equal to + y y = + ) ( ) d y = + = + ( y ) = ( y ) + y ( y ) = ( + y ) ( y ). y = b. If y = log, then d =, so ds d = y log y + y + = log 3 = log3. ( ) = / d =, since is positive on [, ]. the length of the curve is 3 3 d = t t dt = t dt = t log t 3 t = 3 ( ) 3 log 3 where t =. c. If y = log(cos), then = 3 + log ( ( 3 )( ) ), d = tan(), so ds d = + tan () = sec, the length of the curve is 3 3 sec() d = log(sec + tan) = log( + 3). Solution to Eercise 9. The area of R is equal to cos() d = sin ( ) d = cos ( ) =. If R is revolved around the -ais then cross sections perpendicular to the -ais of the resulting solid are circular disks of of radius cos(), for. So the volume of the solid is equal to ( cos()) d =. If R is revolved about the y-ais then the resulting solid consists of concentric cylindrical shells of radius height cos(), for. So the volume of the solid is equal to cos() d = sin ( ) d = cos ( ) + cos ( ) d = 8 sin ( ) = 8. If R is revolved about the line defined by y =, then cross sections perpendicular to the -ais of the generated solid are annuli of inner radius cos() out radius, for. So the volume of the solid is equal to ( ) cos() d = cos() d ( cos()) d = 8.

10 Solutions to Review () Calculus II (-nyb-5/5,6) Winter 9 Solution to Eercise. A four dimensional sphere of radius r consists of sets (,y,z): + y + z r t, for r t r, each of which is a three dimensional sphere of radius r t hence volume 3 (r t ) 3/. Thus, the volume of a four dimensional sphere of radius r is equal to r 3 (r t ) 3/ dt = 3 8 r ( p ) 3/ dp, r by symmetry the change of variables t = pr. Partial integration gives ( p ) 3/ dp = p( p ) 3/ + 3 p ( p ) / dp = 3 (p )( p ) / dp = 3 ( p ) 3/ dp + 3 ( p ) / dp = 3 ( p ) / dp, where the first term on the right in the second line is absorbed by the left h side. Since ( p ) / dp = is one quarter the area of a unit circle, it follows that the volume of a four dimensional sphere of radius r is equal to 8 3 r 3 = r. for all real values of, so there is a real number B such that 5B = e 5 w = e 5, or d d = 5Be5. Thus, there is a real number A such that A + Be 5 = z = e y, or y = Ae + Be 5, as required. Solution to Eercise. a. Separating the variables gives integrating then gives 6 = 6cos()cos() = 3cos(3) + 3cos(), y d log y 6 = sin(3) 3sin() + C, or y 6 = Aep ( sin(3) + 3sin() ). Since y = e is =, e6 = Ae +3, or A = e. Thus, y = ep ( 3 6 sin(3) sin()). b. Separating the variables gives d ( ) dt = te t ( t), integrating (using partial integration on the right side) gives = te t e t t ( t) dt = te t t t + e t + C = e t t + C. If = when t =, then = + C, or C =. = e t ( t), or = e t ( t). Solution to Eercise. First of all d ( e ) = e d (, e ) d d = e, d ( e 3 ) = 3e 3 d d d ( e 3 ) = 9e 3, so e e 3 are solutions of the equation; thus, any sum of multiples of these functions is a solution of the equation (by the linearity of the derivatve). Net, if y is any solution of the equation, z = e y w = d, then w = e ( d + y ) d ( ) e 5 d w = e 3( ) d d d + y = e 3( d y d ) d + 6y =