Triple integral finding bounds?

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beallio
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Homework Statement



Integrate the function over the solid given by the figure below (the bounding shapes are planes perpendicular to the x-y plane, a cone centered about the positive z-axis with vertex at the origin, and a sphere centered at the origin), if P=(0,0,5),Q=(0,4,3), and R=(sqrt(6),sqrt(10),0) .



https://instruct.math.lsa.umich.edu/webwork2_course_files/ma215-f08/tmp/gif/homework8b-prob1-pimages/sfig16-8-1g4.gif


Homework Equations





The Attempt at a Solution


planes perpendicular to the x-y plane are the x-z plane and the y-z plane

how do you use the figure to get the bounds for x,y,z

I think that x is from 0 to R
then y is from 0 to 4
and z is from 0 to 5

is this correct? do I need to switch to spirical coordinates?
 
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beallio said:
Integrate the function over the solid given by the figure below (the bounding shapes are planes perpendicular to the x-y plane, a cone centered about the positive z-axis with vertex at the origin, and a sphere centered at the origin), if P=(0,0,5),Q=(0,4,3), and R=(sqrt(6),sqrt(10),0) .

planes perpendicular to the x-y plane are the x-z plane and the y-z plane

how do you use the figure to get the bounds for x,y,z

I think that x is from 0 to R
then y is from 0 to 4
and z is from 0 to 5

Hi beallio! Welcome to PF! :smile:

No, "planes perpendicular to the x-y plane" means the y-z plane and the vertical plane through R.

The top surface is part of a sphere, and the "front surface" is also curved … it's the cone through Q.

Hint: divide the solid into horizontal slices of thickness dz, and integrate over z (split the integration into two parts … one below Q, and one above Q, since they'll be different functions). :wink: