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Each of the following integrals represents the volume of either a hemisphere or a cone, and the variable of integration measures a length. In each case, say which shape is represented and give the radius of the hemisphere or radius and height of the cone. Make a sketch of the region, showing the slice used to find the integral, labeling the variable and differential on your sketch. Then evaluate the integral to find the area.
A. ∫10 0 π (2 - y/5)^2 dy
1. Which is the shape of the region being integrated?
O Cone
O Hemisphere
2. radius/radius and height = __________ (Enter the radius, or the radius and height separated by a comma, e.g., 4. 3)
3. volume = __________
B. ∫14 0 π(196 - h^2) dh
1. Which is the shape of the region being integrated?
O Cone
O Hemisphere
2. radius/radius and height = __________ (Enter the radius, or the radius and height separated by a comma, e.g., 4, 3)
3. volume = __________

Respuesta :

gkosworldreign

Answer:

1. cone

   radius, height = 2, 10

  volume = 13.33

2. hemisphere

   radius=  14

  volume = 1829.33

Step-by-step explanation:

The  region  described  has  a  radius r=2−y/5, therefore, the volume is a cone

radius, height = 2, 10

volume =

[tex]\int\limits^{10}_0 {\pi ( 2-\frac{y}{5} )^2} \, dy \\=[/tex]

= 13.33

2. The  region  ∫14 0 π(196 - h^2) dh   has  a  radius 196 - h^2,  so the volume is a hemisphere

radius  = 14

volume

=[tex]\int\limits^{14}_0 {\pi (196 - h^{2} )} \, dh[/tex]

=[tex]\pi (196h- \frac{h^3}{3}) |^{14}_0[/tex]

=1829.33

Ver imagen gkosworldreign
Ver imagen gkosworldreign

The correct responses are;

Part A

  1. Cone
  2. Radius and height = 2, 10
  3. Volume [tex]\frac{1}{3} \cdot 40 \cdot \pi[/tex] ≈ 40.89

Part B

  1. Hemisphere
  2. Radius = 14
  3. Volume = [tex]\frac{2}{3}\cdot 2744 \cdot \pi \approx 5747.02 [/tex]

Methods used to obtain the above values

A. The given integral is presented as follows;

[tex]\displaystyle \mathbf{ \int\limits^{10}_0 {\pi \cdot \left(2 - \frac{y}{5} \right)^2} \, dy }[/tex]

The expression, [tex]2 - \dfrac{y}{5} [/tex], is a linear expression and is equivalent to a linear dimension such as a radius, such that we have;

[tex]x = \mathbf{ 2 - \dfrac{y}{5} }[/tex]

At y = 0, the radius of the base is therefore;

[tex]x = r = 2 - \dfrac{0}{5} =2[/tex]

At y = 10, we have;

[tex]x = 2 - \dfrac{10}{5} = 0[/tex]

Which gives a cone with base radius, r = 2, and height where the radius, r  is 0 as h = 10

Therefore;

1. Cone

2. Radius and height = 2, 10

The volume is therefore;

[tex]\displaystyle \int\limits^{10}_0 { \pi \cdot \left(2 - \frac{y}{5} } \right) \, dy = \left[-\frac{1}{75} \cdot \pi \cdot \left(10 - y\right)^3 \right]^{10}_0 = 0 - \left(-\frac{1}{75} \times \pi \times 10^3 \right) = \frac{1}{3} \cdot 40 \cdot \pi[/tex]

3. The volume is; [tex]\underline{\dfrac{1}{3} \cdot 40 \cdot \pi \approx 41.89 } [/tex]

B. The given integral is presented as follows;

[tex]\displaystyle \mathbf{ \int\limits^{14}_0 {\pi \cdot (196 - h^2)} \, dh }[/tex]

The expression, 196 - h², is similar to the expression for the x² value of

the equation of a circle given as follows;

r² = x² + y²

x² = r² - y²

Which gives;

r² = 196 = Constant

r = √(196) = 14

y² = h²

Where;

x = The length of the radius at height, y

Therefore, the given expression represents the equation of a hemisphere.

1. Hemisphere;

2. Radius = 14

The volume is given as follows;

[tex]\displaystyle \int\limits^{14}_0 {\pi \cdot \left(196 - h^2\right)} \, dh = \left[\pi \cdot \left(196 \cdot h - \frac{h^3}{3} \right) \right]^{14}_0 = \pi \times \left(196\times 14 - \frac{14^3}{3} \right) [/tex]

[tex]\pi \cdot \left(196 \times 14 - \dfrac{14^3}{3}\right) = \dfrac{5488\cdot \pi}{3} = \dfrac{2}{3} \cdot 2744\cdot \pi \approx \mathbf{ 5747.02}[/tex]

3. Volume ≈ 5,747.02

Learn more about the derivation of the volume of a solid by integration here:

https://brainly.com/question/2264237

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