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using integrals, find the volume of the solid that lies above the cone z=sqrt(3x

ID: 2979520 • Letter: U

Question

using integrals, find the volume of the solid that lies above the cone z=sqrt(3x^2+3y^2) and below the sphere x^2+y^2+z^2=36

Explanation / Answer

Using spherical coordinates, x^2 + y^2 + z^2 = 36 ==> ? = 6. z = ?(3x^2 + 3y^2) ==> ? cos ? = ?3 * (? sin ?) ==> ? = arctan(1/?3) = ?/6. So, the volume ??? 1 dV equals ?(? = 0 to 2?) ?(? = ?/6 to ?/2) ?(? = 0 to 6) 1 * (?^2 sin ? d? d? d?). (Note that the bounds for ? are due to being below the cone and above the xy-plane.) Evaluating this yields 2? ?(? = ?/6 to ?/2) sin ? d? * ?(? = 0 to 6) ?^2 d? = 2? * -cos ? {for ? = ?/6 to ?/2} * (1/3)?^3 {for ? = 0 to 6} = 2? * ?3/2 * (1/3) 6^3 = 72??3.

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