1. ## spherical coordinates?

Integrate the function
f(x,y,z)=6*x+5*y over the solid given by the "slice" of an ice-cream cone in the first
octant bounded by the planes x=0 and y=sqrt(43/5)*x and contained in a sphere centered at
the origin with radius 13 and a cone opening upwards from the origin with top radius 12."

2. Look at the hand-drawn diagram below.

The diagram on the left is the shape which we are integrating. You can see the sphere of radius 13 outside. And a red cyclinder of radius 12 inside. The blue line is the line drawn from the origin to the place where the cylinder and sphere meet. The angle $\phi$ is the angle made by the z-axis and the blue line. Using some trigonometry (on the triangle formed by the black,red, and blue lines) we find that $\phi = \tan^{-1} (12/5)$.

The diagram to the right is the picture of the region over which we are integrating. The red line is the graph of the line and the angle $\theta$ is the angle between the x-axis and the red line (graph). Which is $\theta = \tan^{-1} (\sqrt{43/5})$.

In spherical coordinate we know that $x = \rho \sin \phi \cos \theta$ and $y = \rho \sin \phi \sin \theta$.

Thus,
$\iint_V 6x+5y dV = \int_0^{\theta} \int_0^{\phi} \int_0^{13} (6\rho \sin \phi \cos \theta + 5\rho \sin \phi \sin \theta) (\rho \sin^2 \phi) \ d\rho \ d\phi \ d\theta$.

3. You don't know how much you've helped, thank you.
Sorry if I irritated you or anyone else, typical newbie move I guess
Ellen

### if a graph of y(x) is transformed in to the graph of 2y-6= -4* f(x)-3

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