the base of a solid is the region bounded by y=squareroot(x), y=2-x and the x-axis. For each y, where 0<=y<=1, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose height is half the length of the base in the region.
The answer is 17/20=.85, but i need the process
The answer is 17/20=.85, but i need the process
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Since we're going "perpendicular to the y-axis", we need "dy" slices. Then
x = 2 - y
and x = y².
The base of each rectangle is b = 2 - y - y²
and the height h = b/2, so the area of each rectangle is
A = b²/2 = (2 - y - y²)²/2 = ½(y^4 + 2y³ - 3y² - 4y + 4). Then
V = ∫ A(y) dy = ½∫[0,1] (y^4 + 2y³ - 3y² - 4y + 4) dy
V = ½((1/5)y^5 + ½y^4 - y³ - 2y² + 4y) |[0,1] = ½(1/5 + ½ - 1 - 2 + 4 - (0))
V = (1/20)(2 + 5 + 10) = 17/20
x = 2 - y
and x = y².
The base of each rectangle is b = 2 - y - y²
and the height h = b/2, so the area of each rectangle is
A = b²/2 = (2 - y - y²)²/2 = ½(y^4 + 2y³ - 3y² - 4y + 4). Then
V = ∫ A(y) dy = ½∫[0,1] (y^4 + 2y³ - 3y² - 4y + 4) dy
V = ½((1/5)y^5 + ½y^4 - y³ - 2y² + 4y) |[0,1] = ½(1/5 + ½ - 1 - 2 + 4 - (0))
V = (1/20)(2 + 5 + 10) = 17/20