volume of the solid obtained by rotating the region

Question

Find the volume of the solid obtained by rotating the region bounded by  , , and  about the -axis.

Answer 1

In this case I'll assume that y = x² is the equation of your curve.

I wish I can draw the figure for you but the features of this tool is very limited. So I'll just solve it for you without the graph.

Using the disk method, we are going to rotate the area around the y-axis. Therefore the equation of the curve should be in terms of y.

x =  √ y

The formula is:

_b

V= ∏ [f(y)]² dy

^a

Since the equation of the curve we need is in terms of y, the values of a and b are from the y-axis.

Using the equation of the curve, If x=0 then y=0, If x=1 then y=1

Therefore the values of a=0 and b=1

Now we can express the area using definite integral.

₁

= ∏[ √y ]² dy

⁰

₁

= ∏ (y²/2)]

0

=(1/2)∏ cubic unit

Answer 2

We can get the volume of the solid by integrating the "dy" from y = 0 to y = 1, that "dy" has a cross section that is a "washer" shape with inside radius of r_i (as defined by the y = x² function) and an outside radius of 1 (as defined by x = 1).

Since the r_i is the "x" in the y = x² function, we can solve y = x² for x getting x = √y. The area of the "washer" is π(r_o² - r_i²) and we would add all the washers stacked up from 0 to 1 with the following integral:

V = ₀∫¹π(r_o² - r_i²)dy

V = ₀∫¹π(1² - (√y)²)dy

V = π•₀∫¹(1 - y)dy

V = π[y - y²/2] evaluated from 0 to 1

V = π[(1 - 1²/2) - (0 - 0²/2)]

V = π(1/2 - 0)

V = 1/2π