# Related Problems

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Q1: Let R be the region enclosed by the curves y=x+2 and y=x^{2}

(a) find the area of R

(b) find the volume of the solid obtained by rotating R about the line y=-1

Answer: (a)The enclosed region is below the curve of y=x+2, and above the curve of y=-1. So the region is in between.

Using integration along x axis,we can get

Area=\int (x+2-x^{2})dx from -1 to 2

=[1/2*x^{2}+2x-1/3*x^{3}] from -1 to 2

=(2+4-8/3)-(1/2-2+1/3)=5-1/2=9/2

(b)We can use disk method to calculate area of each small disk, and then sum them up. For the outer region disk,the radius is x+2+1, for the inner disk, the radius is X^{2}+1. So the area of each disk is the area of the outer disk minus the area of the inner disk.

Volume=\int pi*[(x+2+1)^{2}-(x^{2}+1)^{2}]dx from -1 to 2

=\int pi*(-x^{4}-x^{2}+6x+8)dx from -1 to 2

=pi*[-1/5x^{5}-1/3x^{3}+3x^{2}+8x] from -1 to 2

=pi*(30-33/5)

=117/5*pi

Q2. Find the number b such that the line y=b divides the region bounded by the curves y=x^{2} and y=4 into two regions with equal area.

Answer:We can interpret the question in this way. The region bounded by y=b and y=x^{2} is half of the area of region bounded by y=x^{2} and y=4.

Area of the total region A1

=\int (4-x^{2}) from -2 to 2

=[4x-1/3*x^{3}] from -2 to 2

=16-3/16

=32/3

Curve y=x^{2} intersects y=b with two points (-sqrt(b),b) and (sqrt(b),b)

Area of the sub-region A2

=\int (b-^{2}) from -sqrt(b) to sqrt(b)

=[bx-1/3*x^{3}] from -sqrt(b) to sqrt(b)

=4/3*b^{3/2}

so A1=2*A2

32/3=8/3*b^{3/2}

b=16^{3/2}

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