Calculus 2 Lia Vas Arc Length. Surface Area.

1 Calculus 2 Lia VasArc length . Surface thaty=f(x) is a continuous function with a continuous derivative on[a,b].The arc lengthLoff(x) fora x bcan be obtained by integrating the length length elementdson a sufficiently small interval can be approximated by thehypotenuse of a triangle with +dy2 ds= dx2+dy2and soL= bads= ba dx2+dy2= ba (1 +dy2dx2)dx2= ba (1 +dy2dx2) thatdy2dx2=(dydx)2= (y ) the formula for the arc length becomesL= ba 1 + (y ) of a Surface of thaty=f(x) is a continuous function with acontinuous derivative on [a,b].To compute the Surface areaSxof the Surface obtained by rotatingf(x) aboutx-axis on [a,b],we can integrate the Surface area elementdSwhich can be approximatedas the product of the circumference 2 yof the circle with radiusyand the height that is given bythe arc length 1 + (y )2dx,the formula that computes the Surface area isSx= ba2 y 1 + (y ) (x) is rotated abouty-axis on [a,b],thendSis the product of the circumference 2 xofthe circle with radiusxand the height that is given by the arc length , the formulathat computes the Surface area isSy= ba2 x 1 + (y )

2 Find the length of the curvey=x3/2,1 x Find the length of the curvey= 1 x2, 1 x Use the Left-Right sum calculator program to approximate the length of the curvey=x3,for0 x 1 to two Use the Left-Right sum calculator program to approximate the length of the curvey= sinx,for 0 x to five Use the Left-Right sum calculator program withn= 100 subintervals to approximate thelength of the curvey=ex,for 0 x Find the area of the Surface obtained by rotatingy=x3,for 0 x 2 about Find the area of the Surface obtained by rotatingy= x,for 4 x 9 about Find the area of the Surface obtained by rotatingy=x2,for 1 x 2 about Prove the formula 4r2 computes the Surface area of a sphere with Use the Left-Right sum calculator program to approximate the Surface area obtained by rotatingthe curvey= sinx,for 0 x aboutx-axis to four Use the Left-Right sum calculator program with 100 subintervals to find the Left sum whichapproximates the Surface area of the Surface obtained by rotatingy=ex2+10 x 1, Use the Left-Right sum calculator program with 100 subintervals to find the Right sum whichapproximates the Surface area of the Surface obtained by rotatingy= ln(x3+ 1) 0 x 1, solid with infinite Surface area that encloses a finite Surface of rev-olution obtained by revolvingy=1xfor 1 x is known as theGabriel s Horn orTorricelli s trumpet.

3 Using the inequality 1 +1x4>1,demonstrate that this Surface hasinfinite Surface area . Then find the volume enclosed by this Surface and show it is Calculus 3, we will encounter another example of a similar phenomenon: a fractal objectcalledKoch snowflakewith infinite perimeter that encloses a finite =32x1/2so (y )2=94x L= 41 1 + this integral using the substitutionu= 1 +94xand obtain4923(1 +94x)3/2|41=827(103/2 (134)3/2) = The key step in this problem is to simplify the formula 1 + (y ) derivative isy =12(1 x2) 1/2( 2x) = x 1 x2so 1 + (y )2= 1 +x21 x2=1 x2+x21 x2=11 , the length isL= 1 1 11 x2dx= 1 11 1 x2dx= sin 1x|1 1= sin 1(1) sin 1( 1) = 2+ 2= .3. Careful:firstwrite down the integral that you need to evaluate using the formula for the arclength,thenuse the calculator. Do not enterx3inY1because in that case the program wouldgive you the area under the curve, not the y = the integralL= 10 1 + 9x4dxcomputes the arc length .

4 To evaluatethis integral, enter the function 1 + 9x4asY1in your calculator and use the program for leftand right sums. Withn= 300, you obtain that the length is approximately The problems is asking for thearc lengthnot the area under the curve so, as in the previousproblem, you need to use the formula for the arc length first,beforeentering any function inthe sinx y = cosx. L= 0 1 + the function 1 + cos2xasY1in your calculator and use the program for left and right sums. Withn= 100, you obtainthat the length is approximately y =ex. L= 10 1 + (ex)2dx= 10 1 + 1 +e2xasy1and use theLeft-Right Sums program witha= 0,b= 1 andn= the length of y = 3x2. Sx= 202 y 1 + (y )2dx= 2 20x3 1 + this integralusing the substitutionu= 1 + 2 13623(1 + 9x4)3/2|20= 27(1453/2 1) = x=x1/2 y =12x 1/2=12 x. Sx= 942 y 1 + (y )2dx= 2 94 x 1 + function first. Obtain 2 94 x 4x+14xdx= 2 94 x 4x+12 xdx= 94 4x+ integral usingu= 1 + 1423(1 + 4x)3/2|94= 6(373/2 173/2) = y = 2x.

5 Sy= 212 x 1 + (y )2dx= 2 21x 1 + this integral usingthe substitutionu= 1 + 2 1823(1 + 4x2)3/2|21= 6(173/2 53/2) = You can represent the sphere as the Surface of revolution of the upper part of the circlex2+y2=r2aroundx-axis. So,y= r2 upper half is given by the positive root. Thebounds forxare randr. The derivative isy =12(r2 x2) 1/2( 2x) = x r2 problem 2. in part a), the key step in this problem is to simplify the formula 1 + (y ) + (y )2= 1 +x2r2 x2=r2 x2+x2r2 x2=r2r2 , the Surface area isSx= r r2 y r2r2 x2dx= r r2 r2 x2r r2 x2dx= r r2 rdx= 2 rx|r r= 2 r(r+r) = 4r2 .10. Careful:firstwrite down the integral that you need to evaluate using the formula for the surfacearea,thenuse the calculator. Do not enter sinxinY1because the program would give you thearea under the curve in that case, not the Surface area of the Surface of sinx y = cosx. Sx= 02 sinx 1 + the function 2 sinx 1 + cos2xasY1in your calculator and use the program for left and right sums.

6 Withn= 100,obtainthat the Surface area is approximately The problems is asking for the Surface areaSx= ba2 y 1 + (y )2dx. Find the derivativeof the function and plug it in the formula +1 y =ex2+12x Sx= 102 ex2+1 1 + (2xex2+1) enter 2 ex2+1 1 + (2xex2+1)2asy1(careful with the paren-thesis) and use the program witha= 0,b= 1 andn= that the Surface are isapproximately The problems is asking for the Surface areaSy= ba2 x 1 + (y ) the derivativeof the function and plug it in the formula ln(x3+ 1) y =3x2x3+1 Sx= 102 x 1 + (3x2x3+1) enter 2 x 1 + (3x2x3+1)2or its simplified form 2 x 1 +9x4(x3+1)2asy1(careful with the parenthesis) and use the program witha= 0,b= 1 andn= the Surface are is approximately y = x 2= Surface area isSx= 12 1x 1 + the giveninequality, this integral is larger than 12 1x 1dx= 2 11xdx= 2 lnx| 1= .So, thesurface area is larger than the value of this divergent integral.

7 So,Sxis infinite as , on the other hand, is computed asVx= 1 (1x)2dx= 11x2dx= 1x| 1= ( 1 ( 1)) = .4