Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.
Have an amazing day and I hope this can somewhat help :)
Demire karbon ilavesi ne yapar, ne değiştirir, özellikleri nasıl değiştirir, yansımaları nelerdir? Grafiklerde kullanarak detaylı olarak açıklayın. HMK brainly.com/app/profile/7139574 takip et!!!!
<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form
</h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1)
</h2><h2>With the velocity given by (2.1), the vorticity takes the form
</h2><h2>ω = ∇ × u =
</h2><h2></h2><h2>∂v
</h2><h2>∂x −
</h2><h2>∂u
</h2><h2>∂y
</h2><h2>k. (2.2)
</h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence
</h2><h2>∂v
</h2><h2>∂x −
</h2><h2>∂u
</h2><h2>∂y = 0. (2.3)
</h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity
</h2><h2>potential φ such that u ≡ ∇φ, that is
</h2><h2>u =
</h2><h2>∂φ
</h2><h2>∂x, v =
</h2><h2>∂φ
</h2><h2>∂y . (2.4)
</h2><h2>We also recall the definition of φ as
</h2><h2>φ(x, y, t) = φ0(t) + Z x
</h2><h2>0
</h2><h2>u · dx = φ0(t) + Z x
</h2><h2>0
</h2><h2>(u dx + v dy), (2.5)
</h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent
</h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is
</h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider
</h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane,
</h2><h2>Green’s Theorem implies that
</h2><h2></h2><h2></h2><h2></h2><h2></h2><h2></h2><h2></h2><h2></h2>
