Use a definite integral to derive a formula for the surface area of the indicated solids
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Use a definite integral to derive a formula for the surface area of the indicated solids

[From: ] [author: ] [Date: 12-08-02] [Hit: ]
0)) and substituting it into the point-slope formula,To find dx/dy,Notice that √(h² + r²), by use of the Pythagorean theorem, is equivalent to the length of the slant height. So,......
a) A right circular cone of altitude h and base radius r.
b) A sphere of radius r

Please show the solution step by step, thank you.

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A)
Here's an image of the situation: http://imgur.com/jfAHF

The slant height of a right circular cone with height h and radius r forms a segment of a line that can be described by
y = -h/r x + h

This equation comes from the fact that on the coordinate plane, the slant height connects the points (0, h) and (r, 0). The slope of the line containing these points is
(h - 0)/(0 - r) = -h/r

Then, using one of the points (I use (r, 0)) and substituting it into the point-slope formula, you have
y - 0 = -h/r (x - r)
y = -h/r x + h

Writing this curve as a function of x in terms of y:
x = -r/h y + r
(this form will be used because you're revolving around the y-axis)

Now for the actual integral:
A = 2π ∫ {0 to h} x ds

where
ds = √(1 + (dx/dy)²) dy

A = 2π ∫ {0 to h} (-r/h y + r) √(1 + (dx/dy)²) dy

To find dx/dy, simply derive x(y):
x = -r/h y + r
dx/dy = -r/h

A = 2π ∫ {0 to h} (-r/h y + r) √(1 + (-r/h)²) dy
A = 2π ∫ {0 to h} (-r/h y + r) √(1 + r²/h²) dy
A = 2π ∫ {0 to h} (-r/h y + r) √((h² + r²)/h²) dy
A = 2π √(h² + r²)/h ∫ {0 to h} (-r/h y + r) dy

A = 2π √(h² + r²)/h [(-r/h)/2 y² + ry] {0 to h}
A = 2π √(h² + r²)/h [(-rh²/(2h) + rh) - (0 + 0)]
A = 2π √(h² + r²)/h (-rh²/(2h) + rh)
A = 2π √(h² + r²)/h (-rh/2 + rh)
A = 2π √(h² + r²)/h (rh/2)
A = πr √(h² + r²)

Notice that √(h² + r²), by use of the Pythagorean theorem, is equivalent to the length of the slant height. So, letting s = slant height, you have
A = π rs

= = = = = = = = = = = = = = = = = = =
B)
Here's the equation for a circle with its center at the origin:
x² + y² = r²

However, you only need half of the circle's perimeter when you revolve it to obtain the surface area of the sphere.
y = √(r² - x²)
(I'm keeping it in this form because I'll be revolving the top half of the circle about the x-axis)

A = 2π ∫ {-r to r} y ds

where
ds = √(1 + (dy/dx)²) dx

dy/dx = 1/2 (r² - x²)^(-1/2) (-2x)
dy/dx = -x/√(r² - x²)

Taking advantage of symmetry,
A = 2 ∙ 2π ∫ {0 to r} y √(1 + (dy/dx)²) dx
A = 4π ∫ {0 to r} √(r² - x²) √(1 + ( -x/√(r² - x²) )²) dx
A = 4π ∫ {0 to r} √(r² - x²) √(1 + x²/(r² - x²)) dx
A = 4π ∫ {0 to r} √(r² - x²) √( (r² - x² + x²)/(r² - x²) ) dx
A = 4π ∫ {0 to r} √(r² - x²) √( r²/(r² - x²) ) dx
A = 4π ∫ {0 to r} √(r² - x²) r/√(r² - x²) dx
A = 4π ∫ {0 to r} r dx
A = 4πr ∫ {0 to r} dx

A = 4πr [x] {0 to r}
A = 4πr [r - 0]
A = 4πr²
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