User polar coordinates to find the volume of the solid above the cone z=(x2+y2) and below the sphere x2+y2+z2=1

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Daphne Broadhurst · Accepted Answer

Step 1 Remember that:   Volume under the surface  z=f(x,y)  and above the region D in xy-plane is ∫∫Df(x,y)dA Step 2 In the given problem  z=r22−x2−y2 The following integral gives the required volume which is above  z-axis ∫∫Dr2−x2−y2dA Where D is the region between the circles of radius  r1  and  r2 But this is only half the volume, therefore volume of the solid which remains is   V=2∫∫Dr2−x2−y2dA Step 3 In polar coordinates, we can define the region D as {(r,θ)∈D ∣ r1&amp;lt;r&amp;lt;r2,0&amp;lt;θ&amp;lt;2π} Therefore   V=2∫{0}2π∫{r1}r2r22−r2(r)dr dθ V=4π∫{r1}r2rr22−r2dr Substitute r22−r2=u And −2rdr=du Limits of inregration will change from ∫_r|r2 to ∫r22−r120 V=−2π∫{r22−r12}0u12du V=−2π[23u32]{r22−r12}0

User polar coordinates to find the volume of the solid above the cone z = \sq

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