Find the volume of the solid of cross sections

Question

Find the volume of the solid whose base is the region in the first quadrant bounded by , y=x^2, y=1 , and the y-axis and whose cross-sections perpendicular to the  x axis are squares.

Answer 1

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Cross section perpendicular x-axis and passing trough point (x,x²) has length (1 -x²). Because squre in the section has area (1 - x²)².

Because volume V = ∫₀¹(1 - x²)²dx = ∫₀¹(1 - 2x² + x⁴)dx = (x - 2x²/3 + x⁵/5)₀¹ = 1 - 2/3 + 1/5 = 8/15

Answer: volume V = 8/15

Answer 2

The volume you want is essentially the sum of a bunch of square-shaped "disks" running through the area between the horizontal line y=1 and the parabola y=x^2. Those two functions intersect at the point (1,1), so the limits of integration will be from x=0 to x=1 when we put everything together. The square "area" of each disk is 4(f(x))^2. The area over which you want to integrate is given by f(x)=1-x^2, subtracting the two functions. So you want 4 times the integral, from x=0 to x=1, of (1-x^2)^2, or "the square of 1 minus x squared". Expanding this out gives 1-2x^2+x^4, and the indefinite integral gives x-2/3x^3+1/5x^5. So, for the final stretch, plug "1" into 4(x-2/3x^3+1/5x^5) and simplify. You should get 32/15.