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X=y^2+1 x=3 solution:- R=3-x =3-(y^2+1) V=∫ Π (3-(y^2+1)). Y=x^2-2x Y=4 y=x X^2-2x=x X^2-3x=0 X(x-3)=0 X=0 x=3 R(x)=4-x^2-2x r (x)=4-2x R^2(x)=16-8(x^2-2x)+(x^2-2x)^2 16-8x^2+16x+x^4-4x^3+4x^2 =16-4x^2+16x+x^4-4x^3 r^2(x)=(4-x)^2 =16-8x+x^2.
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Y=x^2-2x • Y=4 • y=x • X^2-2x=x • X^2-3x=0 • X(x-3)=0 • X=0 x=3 • R(x)=4-x^2-2x • r (x)=4-2x • R^2(x)=16-8(x^2-2x)+(x^2-2x)^2 • 16-8x^2+16x+x^4-4x^3+4x^2 • =16-4x^2+16x+x^4-4x^3 • r^2(x)=(4-x)^2 • =16-8x+x^2