CAL3 Consider the solid shaped like an ice cream cone that is bounded by the fun

Question

CAL3

Consider the solid shaped like an ice cream cone that is bounded by the functions z = square root x^2 + y^2 and z = square root 32 - x^2 - y^2. Set up an integral in polar coordinates to find the volume of this ice cream cone. Instructions: Please enter the integrand in the first answer box. typing theta for theta. Depending on the order of integration you choose, enter dr and d theta in either order into the second and third answer boxes with only one dr or d theta in each box. Then, enter the limits of integration and evaluate the integral to find the volume. Consider the solid under the graph of z = e^ -x^2 - y^2 above the disk x^2 + y^2 0 Set up the integral to find the volume of the solid Instructions: Please enter the integrand in the first answer box. typing theta for theta Depending on the order of integration you choose, enter dr and d theta in either order into the second and third answer boxes with only one dr or d theta in each box. Then, enter the limits of integration. Evaluate the integral and find the volume. Your answer will be in terms of a What does the volume approach as a rightarrow infinity?

Explanation / Answer

1)z=(x2+y2)1/2,z=(32-(x2+y2))1/2

x2+y2=r2

(x2+y2)1/2=(32-(x2+y2))1/2

(x2+y2)=(32-(x2+y2))

r2=32-r2

r2=16

r=4

0<=theta<=2pi

integral[0 to 2pi]integral[0 to 4][(32-(r2))1/2-(r2)1/2 ]r dr dtheta

integral[0 to 2pi]integral[0 to 4][r(32-(r2))1/2-(r2)] dr dtheta

integral[0 to 2pi]integral[0 to 4][r(32-(r2))1/2] dr dtheta -integral[0 to 2pi]integral[0 to 4][(r2)] dr dtheta

integral[0 to 2pi] [0 to 4][(-1/3)(32-(r2))3/2] dtheta -integral[0 to 2pi][0 to 4][(1/3)(r3)] dtheta

integral[0 to 2pi] [(-1/3)(32-16)3/2-(-1/3)(32-0)3/2] dtheta -integral[0 to 2pi][(1/3)(43)] dtheta

=[(-1/3)(32-16)3/2-(-1/3)(32-0)3/2]2pi   -[(1/3)(43)]2pi

=(2/3)pi[323/2-163/2-64]

=111

A=0

B=2pi

C=0

D=4

Volume =111

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