Question

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 = __________

Answer 1

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 =

$\int\limits^{10}_0 {\pi ( 2-\frac{y}{5} )^2} \, dy \\=$

= 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

=$\int\limits^{14}_0 {\pi (196 - h^{2} )} \, dh$

=$\pi (196h- \frac{h^3}{3}) |^{14}_0$

=1829.33