Calculus II Test #1 Solutions |

**A/ Evaluate these integrals**

1) | 2) | 3) |

4) | 5) | 6) |

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**B/ Draw a diagram for and solve each of the following:**

1) Find the area of the region bounded by *y* = *x*^{ 2}** ** and ** y = 2 – x^{ 2}**.

.

2) Find the area of the region bounded by the graphs of ** x = 2y** ,

.

3) Find the area of the region bounded by ** x = y^{ 2} – 1** and

to find the points of intersection, set *y*** + 1**** =** *y*^{2} – 1

.

**C/ Draw a diagram for questions 1 and 2 and solve:**

1) Use **Disk Method** to evaluate the volume of the solid obtained by

revolving the region bounded by *y* = *x*** ^{ 2} **and

a) the ** x**-axis

Since *x ^{3} < x^{2}* on the interval (0, 1), the top is

so the outer

.

b) the ** y**-axis if

when we revolve about the ** y**-axis, the outer radius is

2) Use **Shell Method** to find the volume of the solid obtained by

revolving the region between ** x = y^{ 2} **,

Since the shells are standing on the ** x**-axis, we will integrate with respect to

.

3) Find ** z** if

for Definite Integrals if

So, we want **z** such that *f* (*z*)(*b* – *a*) = 36. *f *(*z*) = 3*z*^{2} – 4*z* + 1 and (*b* – *a*) = 4 – 1 = 3.

This means 3 (3*z*^{2} – 4*z* + 1) = 36 so 3*z*^{2} – 4*z* + 1 = 12 therefore 3*z*^{2} – 4*z* – 11 = 0

since we want *z* to be between 1 and 4.