Porisma

Worked solution

Read a selected region before splitting rotation disks

The diagram shows y2=6−xy^2=6-x and y=−xy=-x.

(i) State their intersection coordinates.

(ii) Find the exact area of the finite region between the two graphs.

(iii) The shaded region RR is bounded by the y-axis, the line and the lower part of the parabola. Find the volume when RR rotates through 2π2\pi about the y-axis, to 2 decimal places.

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Parabola x=6−y² and line x=−y intersect at (−3,3) and (2,−2). R is shaded between the y-axis and the nearer positive-x boundary, from y=−sqrt(6) to zero; it changes boundary at (2,−2).
Parabola x=6−y² and line x=−y intersect at (−3,3) and (2,−2). R is shaded between the y-axis and the nearer positive-x boundary, from y=−sqrt(6) to zero; it changes boundary at (2,−2).
Parabola x=6−y² and line x=−y intersect at (−3,3) and (2,−2). R is shaded between the y-axis and the nearer positive-x boundary, from y=−sqrt(6) to zero; it changes boundary at (2,−2).

[2008/RJC/Promo/9(modified)]

CJC · Review problems · Question 12

Answer

Intersections (−3,3)(-3,3), (2,−2)(2,-2); area 125/6125/6; selected-region volume 10.3610.36 cubic units.

(i) Solve the intersection equation

x=−yx=-y gives y2=6+yy^2=6+y, so (y−3)(y+2)=0(y-3)(y+2)=0. The intersections are (−3,3)(-3,3) and (2,−2)(2,-2).

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(ii) Subtract left boundary from right boundary

For −2≤y≤3-2\le y\le3, the right boundary is x=6−y2x=6-y^2 and the left boundary is x=−yx=-y. The area is ∫−23(6−y2+y)dy=[6y−y3/3+y2/2]−23=125/6\int_{-2}^3(6-y^2+y)dy=[6y-y^3/3+y^2/2]_{-2}^3=125/6 square units.

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(iii) Split at the line–curve meeting height

The selected shaded region is not the whole region from (ii). For −6≤y≤−2-\sqrt6\le y\le-2, its radius from the y-axis is 6−y26-y^2; for −2≤y≤0-2\le y\le0, it is −y-y. Thus V=π∫−6−2(6−y2)2dy+π∫−20(−y)2dy=π(2886−656)/15≈10.36V=\pi\int_{-\sqrt6}^{-2}(6-y^2)^2dy+\pi\int_{-2}^0(-y)^2dy=\pi(288\sqrt6-656)/15\approx10.36 cubic units. This reproduces the source split and rounding.

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Common mistakes

The shaded region for the volume is only a small part of the full area between the graphs.

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