Volume of a solid and arc length

lamaclass

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I am not sure how to do these two problems:

1. The volume of a solid is obtained by revolving a region in the first quadrant of the xy-plane about the x-axis. The resulting integral is IN, limits 0 and 3, 2 pi x[sup:4hrw3q8b]2[/sup:4hrw3q8b] dx. Graph the region being rotated.

2. The graph of y = f(x) passes through the origin. The arc length from (0,0) to (x, f(x)) is s(x) = IN, limits 0 and x, = sq. rt. of 1+e[sup:4hrw3q8b]t[/sup:4hrw3q8b]dt. Determine the function f(x).
 
Hello, lamaclass!

2. The graph of y=f(t)\displaystyle y = f(t) passes through the origin.

The arc length from t=0 to t=x is:   s(t)=0x1+etdt\displaystyle \text{The arc length from }t=0\text{ to } t=x\text{ is: }\; s(t) \:=\: \int^x_0 \sqrt{1\,+\,e^t}\,dt

Determine the function, f(t).\displaystyle \text{Determine the function, }f(t).

Formula:     L(t)  =  ab1+[f(t)]2dt\displaystyle \text{Formula: }\;\;L(t) \;=\;\int^b_a\sqrt{1\,+\,\left[f'(t)\right]^2}\,dt


So we have:   [f(t)]2  =  etf(t)=e12t\displaystyle \text{So we have: }\;\left[f'(t)\right]^2 \;=\;e^t \quad\Rightarrow\quad f'(t) \:=\:e^{\frac{1}{2}t}

. . Then:   f(t)  =  2e12t+C\displaystyle \text{Then: }\;f(t) \;=\;2e^{\frac{1}{2}t}+ C


"Through the origin" means:f(0)=0\displaystyle \text{"Through the origin" means:}\:f(0)\,=\,0

. . So we have: 2e0+C=0C=2\displaystyle \text{So we have: }\:2e^0\,+\,C \:=\:0 \quad\Rightarrow\quad C \:=\:-2


\(\displaystyle \text{Therefore: }\;f(t) \;=\;2e^\frac{1}{2}t} - 2\)

 
1) V =  π03[2πx2]2dx = π034π2x4dx = 4π303x4dx\displaystyle 1) \ V \ = \ \ \pi \int_{0}^{3}[2\pi x^{2}]^{2}dx \ = \ \pi \int_{0}^{3}4 \pi^{2} x^{4}dx \ = \ 4\pi^{3}\int_{0}^{3} x^{4}dx

= 4π3x55]03 = 972π35, See graph below.\displaystyle = \ \frac{4\pi^{3}x^{5}}{5}\bigg]_{0}^{3} \ = \ \frac{972\pi^{3}}{5}, \ See \ graph \ below.

[attachment=0:69xav0w8]vwx.jpg[/attachment:69xav0w8]
 

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