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If you draw the bounded region in the x,y-plane, you'll find it to be somewhat ambiguous, but since y = 0 cuts the area between the parabola x = y ² and x = 4 perfectly in half, you can use either the top or bottom half. I'll use the top one, i.e. assume y ≥ 0.
For every x taken from the interval [0, 4], we can get a shell with height √x. The distance from x to the axis of revolution, x = 5, is 5 - x, which corresponds to the radius of the shell. The area of this shell is
2π (radius) (height) = 2π (5 - x) √x
Then the volume of the solid is the sum of infinitely many such shells made at every 0 ≤ x ≤ 4, given by the integral
[tex]\displaystyle 2\pi \int_0^4 (5-x)\sqrt x\,\mathrm dx = 2\pi \int_0^4 \left(5x^{1/2}-x^{3/2}\right)\,\mathrm dx \\\\ = 2\pi \left(\frac{10}3x^{3/2}-\frac25x^{5/2}\right)\bigg|_0^4 \\\\ = \boxed{\frac{416\pi}{15}}[/tex]