Calculus II Assignment # 2

This assignment covers

Area Between Curves

Volume of a Solid of Revolution

Arc Length

Volumes by Slicing

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Questions

1) a) Sketch the region bounded by y = – x², and y = x² – 8 and find the area.

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b) Sketch the region bounded by y² = 4 – x, and x + 2y – 1= 0 and find the area.
Integrate with respect to y.

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2) Sketch the region bounded by the graphs of these equations and find the
volume of the solid generated by revolving the region about the indicated axis.

 a) y = x 4 , y = 0, x = 1 about x-axis. b) y = x 3 + 1, x = 0, y = 2 about y-axis c) y = 4x 2,and 4x + y – 8 = 0, about x-axis. d) y = x 3, x = 2, y = 0 about y-axis

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3) Find the arc length of the graph of (x + 3) 2 = 8 (y – 1) 3 from A( – 2, 3/2) to B(5, 3).

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4) A solid's base is the region in the xy-plane bounded by the graphs of
y 2 = 4x and x = 4. Find the volume of the solid if every cross section by a
plane perpendicular to the x-axis is an isoscele right triangle with
one of the equal sides lying in the base of the solid.

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5) A solid's base is the region in the xy-plane bounded by the graph of
x 2 + y 2 = 16. Find the volume of this solid if every cross section by a
plane perpendicular to the y-axis is a rectangle with height equal to
one third the base. (Draw a sketch)

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6) A solid's base is the region in the xy-plane bounded by the graph of
y = 2x 2 and y = 8. Find the volume of this solid if every cross section by a
plane perpendicular to the x-axis is an equilateral triangle. (Draw a sketch)

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Solutions

1) a) the region bounded by y = - x 2 and y = x 2 – 8
The graphs intersect at (±2, – 4)

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b) The region bounded by y² = 4 - x, and x + 2y - 1= 0.
The graphs intersect at ( – 5, 3) and (3, – 1).

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2)

 a) y = x 4 , y = 0, x = 1 about x-axis.Using disk method, the volume is b) y = x 3 + 1, x = 0, y = 2 about y-axisUsing shell method, the volume is c) y = 4x 2,and 4x + y – 8 = 0, about x-axis. The graphs intersect at (-2, 16) and (1, 4). d) y = x 3, x = 2, y = 0 about y-axis

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3) When we solve (x + 3) 2 = 8 (y – 1) 3 for y, we get f (x) = ½(x + 3) 2 / 3 + 1

The arc length formula is , so we need [f '(x)] ²

Now,

We need 1 + [f '(x)] ², which becomes

Notice that the last denominator is a perfect square, so when we take the root, we get

our integral which is

Using the substitution u = 1 + 9(x + 3) 2 / 3 and du = 6(x + 3) – 1 / 3 dx, we get

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4) y 2 = 4x and x = 4.

The triangle's base is = its height = 2y
Since area of a triangle is ½bh we get A = ½(4y²) = 2y² = 8x

The volume of the solid is

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x 2 + y 2 = 16 is a circle center (0, 0) with radius = 4.

Since area of a rectangle is bh we get A = (2x)(2/3 x) = 4/3 x²

The volume of the solid is

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6) y = 2x 2 and y = 8. Find the volume of this solid if every cross section by a
plane perpendicular to the x-axis is an equilateral triangle.

The line y = 8 and the curve intersect at x = ± 2

The equilateral triangle's base is 8 - y, so we find the height h using
the 30°, 60°, 90° triangle ratios.
This gives us that
Since area of a triangle is ½bh we get

The volume of the solid is

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