The area of a disk of revolution at any x about the x- axis is πy² where y=2x. If we integrate this area on the given range of values of x from x=0 to x=1 , we will get the volume of revolution about the x-axis, which here equals,

$\int\limits^1_0 {(pi)y^{2} } \, dx$

which when evaluated gives 4pi/3.

Now we have to calculate the volume of revolution about the y-axis. For that we have to first see by drawing the diagram that the area of the CD like disk centered about the y-axis for any y, as we rotate the triangular area given in the question would be pi - pi*x². if we integrate this area over the range of value of y that is from y=0 to y=2 , we will obtain the volume of revolution about the y-axis, which is given by,

$\int\limits^2_0 {pi - pix^{2} } \, dy = \int\limits^2_0 {1 - y^{2}/4 } \, dy$

If we just evaluate the integral as usual we will get 4pi/3 again(In the second step i have just replaced x with y/2 as given by the equation of the line), which is the same answer we got for the volume of revolution about the x-axis. Which means that the answer B) is correct.

7 0

3 years ago

Can someone please explain how to do this?(Picture above) I'll give brainliest!

Liula [17]

Answer:

f(1)=4

f(-3)=4

f(5)=-2

f(0)=2

f(-1)=0

f(3)=2

Step-by-step explanation:

Pose f(x) as your x value. Now apply to the graph

5 0

3 years ago

Please hurry! Subtract using the number line. 7−(−1) A number line ranging from negative ten to positive ten, with increments of

Olenka [21]

The answer is -1

Hope this helps.

5 0

3 years ago

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