Using cylindrical shells requires that the representative strip in the region to be parallel to the axis of rotation. So, we will have to integrate with respect to y. This means we will want to start by solving for x in terms of y:
x2 / a2 = (b2 - y2) / b2
x = ± a/b √(b2 - y2)
We then integrate an expression for the lateral surface area of a cylinder, 2πrh, where r is the vertical distance away from the x-axis and h is the width of each strip (i.e. the width of the region). Thus r = y and h = 2x.
V = ∫0b 2πy·2a/b √(b2 - y2) dy
= 4aπ/b ∫0b [ y · √(b2 - y2) ] dy u = b2 - y2
du = -2ydy
= -2aπ/b ∫b^20 [ √u ] du
= -2aπ/b · 2/3u3/2 ]b^20
= 4ab2π / 3

Josh F.
02/20/23

Josh F.
02/20/23
Anonimius ..
I have a question: if I have the expression in terms of x it will give me the upper half of the ellipse, that way can I rotate around the x-axis integrating from -a to a or that would be incorrect? Because your answer is rotating around the y-axis right?02/20/23