A 3-inch by 5-inch index card is rotated around a horizontal line and a vertical line to produce two different solids. Which sol
1 answer:
This question is incomplete, the missing diagram is uploaded along this answer below;
Answer:
horizontal V₁ ( 141.37 in³ ) is less than vertical V₂ ( 235.62 in³ )
Step-by-step explanation:
Given the data in the question and as illustrated in the image below;
from the image,
We find the volume of the solid obtained by rotating the card around a horizontal line;
Volume of the cylinder V₁ = πr²h
given that; radii = 3 in and height = 5 in
we substitute
V₁ = π × (3)² × 5
V₁ = π × 9 × 5
V₁ = 141.37 in³
Next we find the volume of the solid obtained by rotating the card around a vertical line;
Volume of the cylinder V₂ = πr²h
given that; radii = 5 in and height = 3 in
we substitute
V₂ = π × (5)² × 3
V₂ = π × 25 × 3
V₂ = 235.62 in³
Hence horizontal V₁ ( 141.37 in³ ) is less than vertical V₂ ( 235.62 in³ )
You might be interested in
So, we know the center is at -1, -3, hmmm what's the radius anyway?
well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-3~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%20-7%20%26%2C%26%20-5~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B-7-%28-1%29%5D%5E2%2B%5B-5-%28-3%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%28-7%2B1%29%5E2%2B%28-5%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B36%2B4%7D%5Cimplies%20r%3D%5Csqrt%7B40%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-1%7D%7B%20h%7D%2C%5Cstackrel%7B-3%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%5Csqrt%7B40%7D%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-1%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D%28%5Csqrt%7B40%7D%29%5E2%5Cimplies%20%28x%2B1%29%5E2%2B%28y%2B3%29%5E2%3D40)
well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus