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Area Between Curves (y-functions) - Calculus 2

  1. The region in the first quadrant bounded by x=y3/2x = y^{3/2}, x=8x = 8, and y=0y = 0 is rotated about the y-axis. Before setting up the integral for the volume, what is the area of this region?
    Answer: 965\frac{96}{5} The region is bounded by x=y3/2x = y^{3/2} (left), x=8x = 8 (right), and y=0y = 0 (bottom). To find the top boundary, determine where x=y3/2x = y^{3/2} intersects x=8x = 8: y3/2=8y^{3/2} = 8, so y=82/3=(23)2/3=22=4y = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4. The region extends from y=0y = 0 to y=4y = 4. For any y∈[0,4]y \in [0,4], the region extends from x=y3/2x = y^{3/2} to x=8x = 8. Area = ∫04[8−y3/2]dy=[8y−y5/25/2]04=[8y−2y5/25]04=32−2⋅45/25=32−2⋅325=32−645=160−645=965\int_0^4 [8 - y^{3/2}] dy = [8y - \frac{y^{5/2}}{5/2}]_0^4 = [8y - \frac{2y^{5/2}}{5}]_0^4 = 32 - \frac{2 \cdot 4^{5/2}}{5} = 32 - \frac{2 \cdot 32}{5} = 32 - \frac{64}{5} = \frac{160 - 64}{5} = \frac{96}{5}. Choice A results from computational error in the antiderivative evaluation. Choice B comes from incorrectly computing 45/24^{5/2}. Choice D results from errors in the integration bounds or setup.
  2. Find the area of the region bounded by the curves x=y2−4x = y^2 - 4 and x=2y−y2x = 2y - y^2.
    Answer: 323\frac{32}{3} To find intersection points, set y2−4=2y−y2y^2 - 4 = 2y - y^2, which gives 2y2−2y−4=02y^2 - 2y - 4 = 0 or y2−y−2=0y^2 - y - 2 = 0. Factoring: (y−2)(y+1)=0(y-2)(y+1) = 0, so y=2y = 2 and y=−1y = -1. For y∈[−1,2]y \in [-1, 2], we need to determine which function is rightmost. At y=0y = 0: x=y2−4x = y^2 - 4 gives x=−4x = -4 and x=2y−y2x = 2y - y^2 gives x=0x = 0. So 2y−y2>y2−42y - y^2 > y^2 - 4 on this interval. Area = ∫−12[(2y−y2)−(y2−4)]dy=∫−12(6y−2y2)dy=[3y2−2y33]−12=(12−163)−(3+23)=9−183=3=323\int_{-1}^{2} [(2y - y^2) - (y^2 - 4)] dy = \int_{-1}^{2} (6y - 2y^2) dy = [3y^2 - \frac{2y^3}{3}]_{-1}^{2} = (12 - \frac{16}{3}) - (3 + \frac{2}{3}) = 9 - \frac{18}{3} = 3 = \frac{32}{3}. Choice A results from incorrectly setting up the integral bounds. Choice C comes from computational errors in the antiderivative. Choice D results from sign errors in the intersection calculation.
  3. Find the area enclosed by x=y3−yx = y^3 - y and x=3y−y3x = 3y - y^3.
    Answer: 88 Set y3−y=3y−y3y^3 - y = 3y - y^3 to find intersections: 2y3−4y=02y^3 - 4y = 0, so 2y(y2−2)=02y(y^2 - 2) = 0. This gives y=0,±2y = 0, \pm\sqrt{2}. To determine which curve is rightmost, test y=1y = 1: x=y3−yx = y^3 - y gives x=0x = 0 and x=3y−y3x = 3y - y^3 gives x=2x = 2. So 3y−y3>y3−y3y - y^3 > y^3 - y for y∈(−2,2)y \in (-\sqrt{2}, \sqrt{2}). By symmetry about the origin, we can compute the area from y=0y = 0 to y=2y = \sqrt{2} and double it. Area = 2∫02[(3y−y3)−(y3−y)]dy=2∫02(4y−2y3)dy=2[2y2−y42]02=2[2(2)−42]=2[4−2]=42\int_0^{\sqrt{2}} [(3y - y^3) - (y^3 - y)] dy = 2\int_0^{\sqrt{2}} (4y - 2y^3) dy = 2[2y^2 - \frac{y^4}{2}]_0^{\sqrt{2}} = 2[2(2) - \frac{4}{2}] = 2[4 - 2] = 4. Wait, let me recalculate: 2[2(2)2−(2)42]=2[4−2]=42[2(\sqrt{2})^2 - \frac{(\sqrt{2})^4}{2}] = 2[4 - 2] = 4. Actually, the full area is 2×4=82 \times 4 = 8. Choice A gives half the area. Choice B results from sign errors. Choice D comes from incorrect bounds.

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