Desperate college student needs help in double integral

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desperate college student needs help in double integral!!!

here is the problem I couldn't solve, anyone got any idea please help me.
thank you very much in advance

1) use double integrals to derive the given formula for the volume of a right circular cone of radius R and height H. the volume of a cone is given by the formula

(pi*R^2*H)/3
I tried to use polar coordinates, but what is troubling me is that I couldn't get H into the formula.


2) use double integrals to derive the given formula for the volume of a cap of a sphere of radius R and height H where 0<H<R. (the cap of a sphere is the portion of the sphere bounded below by the plane z=R-H and bounded above by the plane z=R). the volume of a cap of a sphere with radius R is given by the formula

( pi*H^2*(3R-H))/3
same problem, I couldn't get H into the formula from double integration


3) use double integrals to derive the given formula for the surface are of a cap of a sphere of radius R and height H where 0<H<R. (the cap of a sphere is the portion of the sphere bounded below by the plane z=R-H and bounded above by the plane z=R). the surface area of a cap of a sphere with radius R is given by the formula

2*pi*R*H
same problem, I couldn't get H into the formula from double integration


4) use a double integral to calculate the area for the region in xy-plane bounded by y=H, y=0, x=0, and the line containing the point (a,0), and (b,H) where a,b,H>0 and b<a.


5) use double integral to calculate the area for the sector in the polar plane bounded by the ray thetha=0 and thetha=R>0 and the circle x^2+y^2=R^2 in the first quadrant.


any help on any problem is deeply appreciated
 
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HallsofIvy
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here is the problem I couldn't solve, anyone got any idea please help me.
thank you very much in advance

1) use double integrals to derive the given formula for the volume of a right circular cone of radius R and height H. the volume of a cone is given by the formula

(pi*R^2*H)/3
I tried to use polar coordinates, but what is troubling me is that I couldn't get H into the formula.
Draw a picture. Put the vertex of the cone at (0,0,0), the axis along the z-axis so that the sides of the cone (you are looking at the cone from the side) are lines passing through (0,0,0) and (R,0,h). What is the equation of that line in x and z? If you use cylindrical coordinates, which I recommend, you can replace x in that by r and use it in 3 dimensions. From that equation you can get z as a function of r involving both R and H.

2) use double integrals to derive the given formula for the volume of a cap of a sphere of radius R and height H where 0<H<R. (the cap of a sphere is the portion of the sphere bounded below by the plane z=R-H and bounded above by the plane z=R). the volume of a cap of a sphere with radius R is given by the formula

( pi*H^2*(3R-H))/3
same problem, I couldn't get H into the formula from double integration
Sounds like you should use spherical coordinates for this. Since [itex]\phi[/itex] measures the angle from the z-axis, at the top [itex]\phi[/itex] is 0. If you draw a picture with a radius out to the line R-H, you should see that it forms a right triangle with hypotenuse R and "near side" R- H. [itex]cos(\phi)= R/(R-H)[/itex]. That gives you one of the limits of integration.


3) use double integrals to derive the given formula for the surface are of a cap of a sphere of radius R and height H where 0<H<R. (the cap of a sphere is the portion of the sphere bounded below by the plane z=R-H and bounded above by the plane z=R). the surface area of a cap of a sphere with radius R is given by the formula

2*pi*R*H
same problem, I couldn't get H into the formula from double integration
Yes! Same problem as (2)!


4) use a double integral to calculate the area for the region in xy-plane bounded by y=H, y=0, x=0, and the line containing the point (a,0), and (b,H) where a,b,H>0 and b<a.
Why is this a problem? What is the equation of the line through (a, 0) and (b, H)? Where does it intersect the other boundaries?


5) use double integral to calculate the area for the sector in the polar plane bounded by the ray thetha=0 and thetha=R>0 and the circle x^2+y^2=R^2 in the first quadrant.
Okay, You know [itex]theta[/itex] goes from 0 to R and r goes from 0 to R. Just do the integral.


any help on any problem is deeply appreciated
 

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