2.4 Arc length of a curve and surface area  (Page 5/8)

$y=5x\text{from}x=0\text{to}x=2$

$y=-\frac{1}{2}x+25\text{from}x=1\text{to}x=4$

$x=4y\text{from}y=\mathrm{-1}\text{to}y=1$

Pick an arbitrary linear function $x=g\left(y\right)$ over any interval of your choice $\left(y_1,y_2\right).$ Determine the length of the function and then prove the length is correct by using geometry.

Find the surface area of the volume generated when the curve $y=\sqrt{x}$ revolves around the $x\text{-axis}$ from $\left(1,1\right)$ to $(4,2),$ as seen here.

Find the surface area of the volume generated when the curve $y=x^2$ revolves around the $y\text{-axis}$ from $(1,1)$ to $(3,9).$

$y=x^{3\text{/}2}$ from $\left(0,0\right)\text{to}\left(1,1\right)$

$y=x^{2\text{/}3}$ from $\left(1,1\right)\text{to}\left(8,4\right)$

$y=\frac{1}{3}{\left(x^2+2\right)}^{3\text{/}2}$ from $x=0\text{to}x=1$

$y=\frac{1}{3}{\left(x^2-2\right)}^{3\text{/}2}$ from $x=2$ to $x=4$

[T] $y=e^x$ on $x=0$ to $x=1$

$y=\frac{x^3}{3}+\frac{1}{4x}$ from $x=1\text{to}x=3$

$y=\frac{x^4}{4}+\frac{1}{8x^2}$ from $x=1\text{to}x=2$

$y=\frac{2x^{3\text{/}2}}{3}-\frac{x^{1\text{/}2}}{2}$ from $x=1\text{to}x=4$

$y=\frac{1}{27}{\left(9x^2+6\right)}^{3\text{/}2}$ from $x=0\text{to}x=2$

[T] $y=\text{sin}x$ on $x=0\text{to}x=\pi$

$y=\frac{5-3x}{4}$ from $y=0$ to $y=4$

$x=\frac{1}{2}\left(e^y+e^{\text{−}y}\right)$ from $y=\mathrm{-1}\text{to}y=1$

$x=5y^{3\text{/}2}$ from $y=0$ to $y=1$

[T] $x=y^2$ from $y=0$ to $y=1$

$x=\sqrt{y}$ from $y=0\text{to}y=1$

$x=\frac{2}{3}{\left(y^2+1\right)}^{3\text{/}2}$ from $y=1$ to $y=3$

[T] $x=\text{tan}y$ from $y=0$ to $y=\frac{3}{4}$

[T] $x={\text{cos}}^{2}y$ from $y=-\frac{\pi }{2}$ to $y=\frac{\pi }{2}$

[T] $x=4^y$ from $y=0\text{to}y=2$

[T] $x=\text{ln}\left(y\right)$ on $y=\frac{1}{e}$ to $y=e$

$y=\sqrt{x}$ from $x=2$ to $x=6$

$y=x^3$ from $x=0$ to $x=1$

$y=7x$ from $x=\mathrm{-1}\text{to}x=1$

[T] $y=\frac{1}{x^2}$ from $x=1\text{to}x=3$

$y=\sqrt{4-x^2}$ from $x=0\text{to}x=2$

$y=\sqrt{4-x^2}$ from $x=\mathrm{-1}\text{to}x=1$

$y=5x$ from $x=1\text{to}x=5$

[T] $y=\text{tan}x$ from $x=-\frac{\pi }{4}\text{to}x=\frac{\pi }{4}$

$y=x^2$ from $x=0\text{to}x=2$

$y=\frac{1}{2}{x}^2+\frac{1}{2}$ from $x=0\text{to}x=1$

$y=x+1$ from $x=0\text{to}x=3$

[T] $y=\frac{1}{x}$ from $x=\frac{1}{2}$ to $x=1$

$y=\sqrt[3]{x}$ from $x=1\text{to}x=27$

[T] $y=3x^4$ from $x=0$ to $x=1$

[T] $y=\frac{1}{\sqrt{x}}$ from $x=1$ to $x=3$

[T] $y=\text{cos}x$ from $x=0$ to $x=\frac{\pi }{2}$

The base of a lamp is constructed by revolving a quarter circle $y=\sqrt{2x-x^2}$ around the $y\text{-axis}$ from $x=1$ to $x=2,$ as seen here. Create an integral for the surface area of this curve and compute it.

A light bulb is a sphere with radius $1\text{/}2$ in. with the bottom sliced off to fit exactly onto a cylinder of radius $1\text{/}4$ in. and length $1\text{/}3$ in., as seen here. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is $1\text{/}4$ in. Find the surface area (not including the top or bottom of the cylinder).

[T] A lampshade is constructed by rotating $y=1\text{/}x$ around the $x\text{-axis}$ from $y=1$ to $y=2,$ as seen here. Determine how much material you would need to construct this lampshade—that is, the surface area—accurate to four decimal places.

[T] An anchor drags behind a boat according to the function $y=24e^{\text{−}x\text{/}2}-24,$ where $y$ represents the depth beneath the boat and $x$ is the horizontal distance of the anchor from the back of the boat. If the anchor is $23$ ft below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places.

[T] You are building a bridge that will span $10$ ft. You intend to add decorative rope in the shape of $y=5\left|\text{sin}\left(\left(x\pi \right)\text{/}5\right)\right|,$ where $x$ is the distance in feet from one end of the bridge. Find out how much rope you need to buy, rounded to the nearest foot.

$y=\text{ln}(\text{sin}x)$ from $x=\pi \text{/}4$ to $x=\left(3\pi \right)\text{/}4.$ ( Hint: Recall trigonometric identities.)

Draw graphs of $y=x^2,$ $y=x^6,$ and $y=x^{10}.$ For $y=x^n,$ as $n$ increases, formulate a prediction on the arc length from $\left(0,0\right)$ to $\left(1,1\right).$ Now, compute the lengths of these three functions and determine whether your prediction is correct.

Compare the lengths of the parabola $x=y^2$ and the line $x=by$ from $\left(0,0\right)\text{to}\left(b^2,b\right)$ as $b$ increases. What do you notice?

Solve for the length of $x=y^2$ from $\left(0,0\right)\text{to}\left(1,1\right).$ Show that $x=\left(1\text{/}2\right)y^2$ from $\left(0,0\right)$ to $\left(2,2\right)$ is twice as long. Graph both functions and explain why this is so.

[T] Which is longer between $\left(1,1\right)$ and $\left(2,1\text{/}2\right)\text{:}$ the hyperbola $y=1\text{/}x$ or the graph of $x+2y=3?$

Explain why the surface area is infinite when $y=1\text{/}x$ is rotated around the $x\text{-axis}$ for $1\le x<\infty ,$ but the volume is finite.