Using double in tegrals to find the area/volume

[zhuyilun]

Homework Statement

1. r=4+3*cos($$\theta$$)
2.bounded by the cylinders x^2+y^2=r^2 and y^2+z^2=r^2
3. the solid enclosed by the parabolic cylinder y=x^2 and the planes z=3y and z=2+y

Homework Equations

The Attempt at a Solution

1. i set up the double integral of r and with r= o to 4+3*cos($$\theta$$), $$\theta$$=0 to 2 pi
2. i set up the double integral of (r^2-y^2)^(1/2) and with x=-(r^2-y^2)^(1/2) to x= (r^2-y^2)^(1/2), y= -r to y=r. and muliply the volume by two due to the symmetricity of the graph
3. i know i should find the volume of one and subtract the volume of the second one. but that's all i know. i don't know how to set up the integrals

can anyone tell me what i did in 1 and 2 is right or not and help me w/ 3? thx

 

[mighty2000]

Hi there,

For part 1, you correctly set up the double integral with the limits of integration for r and θ. However, make sure to also include the function that you are integrating, which would be r in this case. So the integral should be ∫∫r dA, where dA is the area element in polar coordinates.

For part 2, you are correct in setting up the integral for the volume of the solid. However, since the solid is symmetric about the x-axis, you only need to integrate from 0 to r, and then multiply the result by 2.

For part 3, you can set up the integral for the first volume as ∫∫y dA, where y is the function y=x^2 and dA is the area element in the xz-plane (since the solid is bounded by the given parabolic cylinder and planes). The limits of integration for x and z would be determined by the intersection of the parabolic cylinder and the planes. Similarly, for the second volume, you can set up the integral as ∫∫(2+y-3y) dA, with the same limits of integration. Then you can subtract the second volume from the first to get the final answer.

Hope this helps! Let me know if you have any other questions.