Arc Length and Surface Area Assignment

Question 1

Approximate the length of $f(x)=x^4$ from $x=0$ to $x=1$ using the following steps:

Divide the interval $[0,1]$ into $5$ equal subdivisions of width $\Delta x=\frac{1}{5}$ .
Calculate the length of the line segment from $(x_i,f(x_i))$ to $(x_{i+1},f(x_{i+1}))$ using the distance formula for $i=0,\ 1,\ 2,\ 3,\ 4$ .
Add up the five lengths from the last step. This is your approximation. Convert your approximation to a decimal using N(_).
Now use numerical_integral to calculate the actual arc length.
Calculate the difference between your approximation and the exact value (this is your error). [Answer: 0.0059]

sqrt((1/5)^2+((1/5)^4)^2)+sqrt((1/5)^2+((2/5)^4-(1/5)^4)^2)+sqrt((1/5)^2+((3/5)^4-(2/5)^4)^2)+sqrt((1/5)^2+((4/5)^4-(3/5)^4)^2)+sqrt((1/5)^2+(1-(4/5)^4)^2)
N(_)

1/625*sqrt(151786) + 1/625*sqrt(15626) + 1/125*sqrt(794) + 1/125*sqrt(634) + 1/25*sqrt(74)
1.59431387309517

integral(sqrt(1+(4*x^3)^2),0,1)
N(_)
1.6002294276722058-1.59431387309517

integrate(sqrt(16*x^6 + 1), x, 0, 1)
1.6002294276722058
0.00591555457703574

Question 2

Find the length of $f(x)=\sqrt{1-x^2}$ from $x=-1$ to $x=1$ . [Answer: $\pi$ ]

Practice your $\LaTeX$ : Write out the integral you used to compute the answer.

derivative(sqrt(1-x^2))

-x/sqrt(-x^2 + 1)

integral(sqrt(1+(-x/sqrt(-x^2 + 1))^2),-1,1)

pi

%md
$\int_{-1}^1\sqrt{1+(\frac{-x}{\sqrt{1-x^2}})}$

$\int_{-1}^1\sqrt{1+(\frac{-x}{\sqrt{1-x^2}})}$

Question 3

Find the area of the surface formed by rotating around the $x$ -axis the graph of $\displaystyle f(x)=\sin(x)$ from $x=0$ to $x=\pi$ . [Answer: 14.42]

Practice your $\LaTeX$ : Write out the integral you used to compute the surface area.

f(x)=sin(x)

integral(2*pi*f(x)*(sqrt(1+cos(x)^2)),0,pi)
N(_)

2*pi*(sqrt(2) + arcsinh(1))
14.4235994484141

%md
$\int_0^\pi2\pi f(x)\sqrt{1+cos(x)^2}$

$\int_0^\pi2\pi f(x)\sqrt{1+cos(x)^2}$

Question 4

Find the area of the surface formed by rotating around the $y$ -axis the graph of $\displaystyle g(y)=y^3$ from $y=1$ to $y=2$ . [Answer: 199.48]

Practice your $\LaTeX$ : Write out the integral you used to compute the surface area.

%var y
g(y)=y^3

integral(2*pi*g(y)*(sqrt(1+(3*y^2)^2)),1,2)
N(_)

1/27*pi*(145*sqrt(145) - 10*sqrt(10))
199.480479701738

%md
$\int_1^22\pi g(y)\sqrt{1+(3y^2)^2}$

$\int_1^22\pi g(y)\sqrt{1+(3y^2)^2}$