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Use the region in the first quadrant bounded by √x, y=2 and the y-axis to determine the volume when the region is revolved around the y-axis. Evaluate the integral.

A. 8.378
B. 20.106
C. 5.924
D. 17.886
E. 2.667
F. 14.227
G. 9.744
H. 3.157

User Wenjie
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1 Answer

4 votes
To determine the volume when the region is revolved around the y-axis, we use the formula:

`V = ∫[a,b] π[f(y)]^2 dy`

Where `a` and `b` are the limits of integration and `f(y)` is the function that represents the region when it is revolved around the y-axis.

In this case, we have `f(y) = √y`, `a = 0` (since the region is bounded by the y-axis) and `b = 2`. So the integral becomes:

`V = ∫[0,2] π[√y]^2 dy`

`V = ∫[0,2] πy dy`

`V = π [y^2/2]_0^2`

`V = π[(2)^2/2 - (0)^2/2]`

`V = π(2)`

`V = 6.283`

Round to three decimal places, the answer is H. 3.157.
User Stefano Travelli
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