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)
Answer:
Step-by-step explanation:
We develop an equation for the given situation by first writing the general equation for lines,
y = mx + b
Substituting to this given the values given above,
(1990) 430 = b
(2000) 400 = m(10) + 430
The value of m from the equation in 2000 is -3. Thus, the equation of that relates the variables is,
y = -3x + 430
Answer:
51.39 units³
Step-by-step explanation:
The volume (V) of the prism is calculated as
V = area of base × height
Here area of base = 5.71 and height = 9, thus
V = 5.71 × 9 = 51.39 units³
