MvCalc43

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The points of intersection of the two curves y=4-x^{2},y=3x\, are (-4,-12),(1,3)\,

The projection R of z=x+4 in the xy-plane is bounded by these curves. So R can be covered by varying y from3x to 4-x^{2}\,(top curve) and x from -4 to 1. Thus the required volume V is given by

V=\int _{{-4}}^{{1}}\int _{{3x}}^{{4-x^{2}}}(x+4)dydx\,

=\int _{{-4}}^{{1}}(x+4)(4-x^{2}-3x)dx\,

=\int _{{-4}}^{{1}}(-x^{3}-7x^{2}-8x+16)dx\,

=[{\frac  {-x^{4}}{4}}-{\frac  {7x^{3}}{3}}-{\frac  {8x^{2}}{2}}+16x]_{{-4}}^{{1}}={\frac  {625}{12}}\,

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