Calculus study realm

SPIRIT?!!@#%^^&*^%@#$ OMG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I'LL NEVER LET YOU GO FOR THE REST OF MY LIFE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

This was his idea. He's completely fine. After twelve hours of check up and recovery, we went back to our apartment and he got idea to do this.

I was fooled by some ghost monkey and its loyal dog.

Dog?

You don't know? "Shadow" is a dog's name.

Ohoho. But thanks for everyone's support. I'll be healed by no time!

You're still injured? Is it serious?

He is severely injure on the buttocks. But he'll be able to walk like normal human being in no time. Thus, nothing will prevent him to do the exam.

XD Oh, I want to see how he walks now. Mehe

Ok, Spirit. Since you're fine, can you do #2,f on the volume of revolution exercise sheet?

Yes, my lady!

Here goes again. It's so late. Don't you want to rest and leave it for tomorrow?

It must be their ritual. XD

I'm going to sleep. See you guys tomorrow! Good night.

Use the washers method. as Dark has explained earlier

π∫(R^2 - r^2) dt = π∫(-1 to 2) ((2 - (y^2 - 2))^2 - (2 - y)^2) dy = π∫(-1 to 2) ((4 - y^2)^2 - (2 - y)^2) dy = π∫(-1 to 2) (16 - 8y^2 + y^4 - (4 - 4y + y^2)) dy = π∫(-1 to 2) (12 - 9y^2 + y^4 + 4y) dy = π(12y - 3y^3 + 1/5 y^5 + 2y^2)|(-1 to 2) = π(12(2) - 3(2)^3 + 1/5 (2)^5 + 2(2)^2 - (12(-1) - 3(-1)^3 + 1/5 (-1)^5 + 2(-1)^2)) = 108/5 π

Can you do this problem: Find the area enclosed by the function y^2 = x^2 - x^4.

To visualize the function:
Function has symmetry in the y-axis, and x-axis (even power)
y^2 = x^2 - x^4 = x^2 (1 - x^2), thus x-intercepts at x = 0, x = ±1
y-intercept at y^2 = (0)^2 - (0)^4 = 0



area = 4∫y dx = 4∫(0 to 1) (x^2 - x^4)^(1/2) dx = 2∫(0 to 1) 2x(1 - x^2)^(1/2) dx = -4/3 (1 - x^2)^(3/2) |(0 to 1) dx = 0 - (-4/3) = 4/3

Find the area of a circle of radius r.

x^2 + y^2 = r^2

area = 4∫(0 to r) dx = 4∫(0 to r)(r^2 - x^2)^(1/2) dx =4∫(0 to π/2) (r^2 - r^2 (sin☼)^2)^(1/2) r cos☼ d☼ = 4r^2 ∫(0 to π/2) (cos☼)^2 d☼ = 4r^2 ∫(0 to π/2) (1/2 + 1/2 cos(2☼)) d☼ = 4r^2 (☼/2 + sin(2☼)/4)|(0 to π/2) = 4r^2 (π/4 + 0) = πr^2
Let x = r sin☼
dx = r cos☼ d☼

Can you use the parametric equation for the last question?

y = r sin♪
x = r cos♪
0 ≤ ♪ ≤ 2π

∫(0 to r) r sin♪ dx = ∫(2π to 0) r sin♪ (r -sin♪) d♪ = r^2 ∫(0 to 2π) (sin♪)^2 d♪ = r^2 ∫(0 to 2π) (1/2 - 1/2cos(2♪)) d♪ = r^2 (♪/2 - sin(2♪)/4)|(0 to 2π) = πr^2

Find the volume generated by the cycloid x = █ - sin█, y = 1 - cos█, where 0 ≤ █ ≤ 2π, and the x-axis about the y-axis.

NB: ignore a

Using shells:
2π∫(0 to 2π) (1 - cos█)x dx = 2π∫(0 to 2π) (1 - cos█)(█ - sin█)(1 - cos█) d█ = 2π∫(0 to 2π) (1 - 2cos█ + (cos█)^2)(█ - sin█) d█ = 2π∫(0 to 2π) (█ - 2█cos█ + █(cos█)^2 - sin█ + 2cos█sin█ - sin█(cos█)^2) d█ = 2π(█^2 /2 + cos█ + (sin█)^2 + (cos█)^3 /3)|(0 to 2π) + 2π∫(0 to 2π) (-2█cos█ + █(cos█)^2) d█ = 2π((2π)^2 /2 + cos(2π) + (cos(2π))^3 /3 - cos(0) - (cos(0))^3 /3) + 2π∫(0 to 2π) (-2█cos█ + █(cos█)^2) d█ = 2π(2π^2 + 1 + 1/3 - 1 - 1/3) + 2π∫(0 to 2π) (-2█cos█ + █(1/2 + 1/2 cos(2█))) d█ = 4π^3 - 4π∫(0 to 2π) █cos█ d█ + 2π∫(0 to 2π) (█/2 + █/2 cos(2█)) d█ = 4π^3 + 2π(█^2 /4)|(0 to 2π) - 4π∫(0 to 2π) █cos█ d█ - 2π∫(0 to 2π) █/2 cos(2█) d█ = 4π^3 + 2π^3 - 4π∫(0 to 2π) █cos█ d█ - π∫(0 to 2π) █cos(2█ d█ = 4π^3 + 2π^3 + 4π█sin█|(0 to 2π) - 4π∫(0 to 2π) sin█ d█ + sin(2█) π█/2|(0 to 2π) - π/2 ∫(0 to 2π) sin(2█) d█ = 4π^3 + 2π^3 + 4π cos█|(0 to 2π) + π/4 cos(2█)|(0 to 2π) = 6n^3

Integrate by parts:
f(x) = █
f'(x) = 1

g(x) = -sin█
g'(x) = cos█

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f(x) = █
f'(x) = 1

g(x) = -sin(2█)/2
g'(x) = cos(2█)

Oh, non. This is too confusing to read...

That's your problem.