The second plane slices the cylinder at an angle... first thing to do is work out the highest and lowest points of this slice .

You'll find these points by using the line x + z = 5 along the diameter of the cylinder, y=0. You are able to choose the value since the plane is independent of y. So solve simultaneously for that line and the cylinder at y=0).

Once you have the heights (z-coordinates) of these points, the volume is simply the cylinder to the lower point plus half the remaining portion (since by looking in the y direction it is clear you have half this last bit of cylinder: one triangle from a rectangle).

Hope this makes things clearer.

Reply 7

ztibor

Original post by Visa Electron

No, it's not a cylinder.

The volume bellow the z=f(x,y) and above the area of A on the z=0 plane

\int \int_A f(x,y)\ dA

and

dA=dx \cdot dy

Your A is the base of the cylinder that is

x^2+y^2=9 \rightarrow y=\sqrt{9-x^2}

As z+x=5, your f(x,y)=-x+5
So

V=\int^3_0 \int^{\sqrt{9-x^2}}_0 (-x+5)\ dy\ dx

But it is better to calculate with polar coordinates

x=r\cdot cos \phi

and

y=r\cdot sin \phi

dA = r\cdot dr \cdot d \phi

and limits of

\phi

0 and

2\pi

(edited 13 years ago)

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