Since the rotation is about the y-axis, I'll integrate by dy.
![\displaystyle y=x^3\\x=\sqrt[3]y\\\\V=\pi \int \limits_1^8(2^2-(\sqrt[3]y)^2)\, dy\\V=\pi \Big[4x-\dfrac{3}{5}x^{\tfrac{5}{3}}\Big]_1^8\\V=\pi \left(4\cdot8-\dfrac{3}{5}\cdot8^{\tfrac{5}{3}\right-\left(4\cdot1-\dfrac{3}{5}\cdot1^{\tfrac{5}{3}\right)\right)\\V=\pi \left(32-\dfrac{96}{5}-\left(4-\dfrac{3}{5}\right)\right)\\V=\pi \left(\dfrac{64}{5}-\dfrac{17}{5}\right)\\V=\dfrac{47\pi}{5}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%3Dx%5E3%5C%5Cx%3D%5Csqrt%5B3%5Dy%5C%5C%5C%5CV%3D%5Cpi%20%5Cint%20%5Climits_1%5E8%282%5E2-%28%5Csqrt%5B3%5Dy%29%5E2%29%5C%2C%20dy%5C%5CV%3D%5Cpi%20%5CBig%5B4x-%5Cdfrac%7B3%7D%7B5%7Dx%5E%7B%5Ctfrac%7B5%7D%7B3%7D%7D%5CBig%5D_1%5E8%5C%5CV%3D%5Cpi%20%5Cleft%284%5Ccdot8-%5Cdfrac%7B3%7D%7B5%7D%5Ccdot8%5E%7B%5Ctfrac%7B5%7D%7B3%7D%5Cright-%5Cleft%284%5Ccdot1-%5Cdfrac%7B3%7D%7B5%7D%5Ccdot1%5E%7B%5Ctfrac%7B5%7D%7B3%7D%5Cright%29%5Cright%29%5C%5CV%3D%5Cpi%20%5Cleft%2832-%5Cdfrac%7B96%7D%7B5%7D-%5Cleft%284-%5Cdfrac%7B3%7D%7B5%7D%5Cright%29%5Cright%29%5C%5CV%3D%5Cpi%20%5Cleft%28%5Cdfrac%7B64%7D%7B5%7D-%5Cdfrac%7B17%7D%7B5%7D%5Cright%29%5C%5CV%3D%5Cdfrac%7B47%5Cpi%7D%7B5%7D%20)
(5/3)^-4 * (5/3)^-5 = (5/3)^(-5-4) = (5/3)^-9
(5/3)^-9 = (5/3)^3x
So -9 = 3x
x = -3 answer
Use the chain rule.
Let u = 25sin²(x), such that dy/dx = dy/du · du/dx


Answer:
Step-by-step explanation:
base=42 m
height=h=?
base angle θ=60°
tan 60=h/42
h=42tan 60=42√3≈72.75 m
height of taller building=135+42√3 m≈135+72.75≈207.75 m≈208 m
The answers C because it is below sea level and sea level is 0 so it has to be negative