Yeah I know why ?????,????
Answer:
a) ⅓ units²
b) 4/15 pi units³
c) 2/3 pi units³
Step-by-step explanation:
4y = x²
2y = x
4y = (2y)²
4y = 4y²
4y² - 4y = 0
y(y-1) = 0
y = 0, 1
x = 0, 2
Area
Integrate: x²/4 - x/2
From 0 to 2
(x³/12 - x²/4)
(8/12 - 4/4) - 0
= -⅓
Area = ⅓
Volume:
Squares and then integrate
Integrate: [x²/4]² - [x/2]²
Integrate: x⁴/16 - x²/4
x⁵/80 - x³/12
Limits 0 to 2
(2⁵/80 - 2³/12) - 0
-4/15
Volume = 4/15 pi
About the x-axis
x² = 4y
x² = 4y²
Integrate the difference
Integrate: 4y² - 4y
4y³/3 - 2y²
Limits 0 to 1
(4/3 - 2) - 0
-2/3
Volume = ⅔ pi
A cone with the same radius and height as a cylinder will have one-third the volume of the cylinder.
Vcone = (1/3)*π*r^2*h
Vcylinder = π*r^2*h
Answer:
x = 8
Step-by-step explanation:
Firstly, create an equation or formula
- because they are congurent, IN = AT, so:
2x+10 = 26 (minus 10 from both sides)
2x = 16 (divide both sides by 2)
x = 8
Sub into the formula:
2 x 8 + 10 = 26
26 = 26
Answer:
a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Step-by-step explanation:
for the equation
(1 + x⁴) dy + x*(1 + 4y²) dx = 0
(1 + x⁴) dy = - x*(1 + 4y²) dx
[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx
∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx
now to solve each integral
I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁
I₂= ∫[-x/(1 + x⁴)] dx
for u= x² → du=x*dx
I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂
then
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C
for y(x=1) = 0
1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C
since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be
1/2 * 0 = - π/4 + C
C= π/4
therefore
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4