So what you'll do is multiply the length times the width and add the height to get your answer.
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All cross sections of spheres are “circles”
If you wanted to calculate it, there it is:
1.) A = πr^2
Rate of change (dA/dr) = 2πr.
When r = 3, dA/dr = 2(3)π = 6π ft^2/ft
2.) V = πr^2h
Rate of change of volume given constant r is dV/dt = πr^2dh/dt = π(6)^2(3) = 108π cm^3/min
The two boundary curves y = √(6x + 4) and y = 2x meet at
√(6x + 4) = 2x
6x + 4 = 4x²
2x² - 3x - 2 = 0
(x - 2) (2x + 1) = 0
⇒ x = -1/2 and x = 2
R is bounded to the left by the y-axis (x = 0), so R is the set
R = {(x, y) : 0 ≤ x ≤ 2 and 2x ≤ y ≤ √(6x + 4)}
Using the shell method, the volume is made up of cylindrical shells of radius x and height √(6x + 4) - 2x. So each shell of thickness ∆x contributes a volume of
2π (radius) (height) ∆x = 2π x (√(6x + 4) - 2x) ∆x
and as we let ∆x approach zero, the total volume of the solid is given by the definite integral
