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Solved: Slicing a hemispherical cake A cake is shaped like
Chapter 13, Problem 62(choose chapter or problem)
Slicing a hemispherical cake A cake is shaped like a hemisphere of radius 4 with its base on the xy-plane. A wedge of the cake is removed by making two slices from the center of the cake outward, perpendicular to the xy-plane and separated by an angle of \(\varphi\).
a. Use a double integral to find the volume of the slice for \(\varphi=\pi / 4\). Use geometry to check your answer.
b. Now suppose the cake is sliced by a plane perpendicular to the xy-plane at \(x=a>0\). Let D be the smaller of the two pieces produced. For what value of a is the volume of D equal to the volume in part (a)?
Questions & Answers
QUESTION:
Slicing a hemispherical cake A cake is shaped like a hemisphere of radius 4 with its base on the xy-plane. A wedge of the cake is removed by making two slices from the center of the cake outward, perpendicular to the xy-plane and separated by an angle of \(\varphi\).
a. Use a double integral to find the volume of the slice for \(\varphi=\pi / 4\). Use geometry to check your answer.
b. Now suppose the cake is sliced by a plane perpendicular to the xy-plane at \(x=a>0\). Let D be the smaller of the two pieces produced. For what value of a is the volume of D equal to the volume in part (a)?
ANSWER:Step 1 of 8
The following are given by the question:
The radius of the hemisphere,\(=4\)
The volume of the slice needs to calculate,\(\phi=\frac{\pi}{4}\)
The coordinates in the polar form are given by,
\(\begin{array}{l} x=r \cos \phi \\ y=r \sin \phi \\ d x d y=r d r d \phi \end{array}\)
The equation of the hemisphere is given by
\(x^{2}+y^{2}+z^{2}=4^{2}\)
Substitute the values and solve as
\(\begin{aligned} (r \cos \phi)^{2}+(r \sin \phi)^{2}+z^{2} & =4^{2} \\ r^{2} \cos ^{2} \phi+r^{2} \sin ^{2} \phi+z^{2} & =16 \\ r^{2}\left(\cos ^{2} \phi+\sin ^{2} \phi\right)+z^{2} & =16 \\ r^{2}+z^{2} & =16 \\ z & =\sqrt{16-r^{2}} \end{aligned}\)