Which of the following integrals correctly computes the volume formed when the region bounded by the curves x^2 + y^2 = 25, x = 3, and y = 0 is rotated around the y-axis?

Question

Question options:

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2)
3)
4)

Answer 1

Answer #2 is the right one.

Pretend for a moment that we're working on a different problem: namely, take the area between the lines y=0, y=4, x=0 and the circle x² + y² = 25. To make it unambiguous, x=0 is the left boundary of this region. Note that the line y=4 intersects the circle at the point (3,4). The volume generated by rotating this region about the y-axis is ∫πr²dy (as y goes from 0 to 4) because the cross sections are disks. The radius is √25 - y² (i.e., for each y, the radius is the x value for which (x,y) is on the circle). For the modified problem, V = ∫π(√25 - y²)²dy; lower limit of integration is y=0 and upper limit is y=4.

Returning to the original problem: now remove the portion of the area above for which x < 3. This corresponds to "boring out" the portion of the solid for which |x| < 3. The volume of the bored-out portion is π*3².