Can someone help me with this? Please? This is not a test/quiz/exam. This is from hw that I do not know how to anwser.

Question

1. The base of a solid in the xy-plane is the circle x² + y² = 16. Cross sections of the solid perpendicular to the y-axis are squares. What is the volume, in cubic units, of the solid? (5 points)

a
b	1024Ï
c
d

1. For an object whose velocity in ft/sec is given by v(t) = -2t² + 8, what is its distance travelled, in feet, on the interval t = 0 to t = 3 secs? (5 points)

a	6.00
b	0.67
c	15.33
d	5.11

1. Which of the following integrals will find the volume of the solid that is formed when the region bounded by the graphs of y = e^x, x = 1, and y = 1 is revolved around the line y = -2. (5 points)

a
b
c
d

Answer 1

Hi Hoshi,

Q1. The first thing you need to do is to determine the length of a side of the cross-sectional squares.

If you solve the given equation for y, you will get

x² + y² = 16

x² = 16 - y²

x = ±√(16 - y²) ⇐ "+" represents the top half and "-" represents the bottom half.

Now, a side of a cross-sectional square can be defined as

(x value on the top half of the circle) - (x value on the bottom of the circle) at a given y.

That is √(16 - y²) - ( - √(16 - y²)) = √(16 - y²) + √(16 - y²) = 2√(16 - y²).

Based on this finding, we can also find the area of each square. ⇒ [2√(16 - y²)]² = 4 * (16 - y²)

Now, take the definite integral between y = - 4 and 4.

4 * ∫_-4⁴ 16 - y² dy. Which will give you the answer. (choice d)

Q2.

This problem is asking you to find the distance it travelled given the velocity.

Here is a bit of physics background knowledge; velocity is defined as the derivative of position.

Hence, the distance it travelled = ∫₀³ v(t) dt = ∫₀³ -2t² +8 dt = 6 (choice a)

Q3. Make sure you draw out the figure.

First, find the outer radius and the inner radius of the shape.

We know that y = e^x > 1 for the interval that we are using.

Let's denote the outer radius as r_o and the inner radius as r_i.

Then r_o = e^x + 3 (notice that 3 is the distance between the axis we are revolving around and the shape)

r_i = 3

First, define the interval that you are taking integral over. ⇒ x = [0,1]

Second, find the volume of the shapes using the outer radius and inner radius separately.π

π∫₀¹ ( e^x + 3 )² dx and π∫₀¹ 3² dx, respectively.

Lastly, subtract the inner figure from the outer figure.

π∫₀¹ (( e^x + 3 )² - 3²) dx (choice d)

Answer 2

hello,

First question has a problem, probably those given answers are wrong

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Second question: first of all find the distance

d = ∫ v(t) dt = ∫ ( -2t^2 + 8 ) dt = -2/3 t^3 + 8 t when bounded t = 0 to t= 3

d = -2/3 ( 3)^3 + 8(3) = -18 + 24 = 6

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question #3

bounded by the graphs of y = e^x, x = 1, and y = 1 is revolved around the line y = -2.

first of all graph y = e^x , x=1 y = 1 and recognize the region

second, choose a point on the graph such A : ( x , e^x ) another point such as B: (1 , 1 )

third, draw y = -2 and make perpendicular from A and B to the y = -2 and named H

v = pi∫ [( HA ) ^ 2 - (HB) ^2 ]dx = ∫[(e^x -(-2))^2 - (1- (-2))^2 ] dx = ∫ [( e^x + 2 ) ^2 - 3^2] dx and bounded of integral will be x = 0 to x = 1

I hope it is useful,

Minoo