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 about ^{ 4} , y = 0, x = 1x-axis. |
b) y = x2 about ^{ 3} + 1, x = 0, y = y-axis |

c) y = 4xand ^{ 2}, 4x + y – 8 = 0, about x-axis. |
d) y = x0 about ^{ 3}, x = 2, y = 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

plane perpendicular to the

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

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

plane perpendicular to the

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**Solutions**

1) a) the region bounded by **y = - x ^{ 2}** and

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 about ^{ 4} , y = 0, x = 1x-axis.Using disk method, the volume is |
b) y = x2 about ^{ 3} + 1, x = 0, y = y-axisUsing shell method, the volume is |

c) y = 4xand ^{ 2}, 4x + y – 8 = 0, about x-axis.The graphs intersect at (-2, 16) and (1, 4). |
d) y = x0 about ^{ 3}, x = 2, y = 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

The triangle's base is = its height = *2y*

Since area of a triangle is ½*bh* 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

plane perpendicular to the

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 ½*bh* we get

The volume of the solid is

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