double integral volume - wedge of cylinder

[sambellamy]

I have been over this problem a bunch of times and can't get it. can someone show me where I'm going wrong?

The problem asks, " find the volume of the solid... the wedge cut from the cylinder 4x² + y² = 9 by the planes z = 0 and z = y + 3

I started with the z = 0 base, and came up with z = 9 - 4x² - y² , which I set to be the f(x,y) in the middle of my integral statement. should I have incorporated the y + 3 here? where does that come into play? I have my integral set up to do type II (dA = dx dy) since I couldn't avoid a trig substitution otherwise, though maybe that was due to an earlier error. anyway so I have the limits of the x (inside) integral being x = (3/2 + y/16) for the upper limit and -(3/2 + y/16) for the lower limit, which I deduced from the ellipse on the xy plane (when z = 0).

I first integrated 9 - 4x² - y² with respect to x with these limits, and got:
-8/3 (3/2 + y/16)³ - 3y² - y³/8.

I then integrated this with respect to y, the y limits being -3 and 3 (again from the ellipse in the xy plane). this somehow ended up in me getting 1755/32 for the volume. this is not correct.

The first thing I tried integrating as my f(x,y) was y + 3. was this the right way to go?

Should I maintain the limits I had? One iteration had me using the y limits as sqrt(9-4x²) and -sqrt(9-4x²), and then the x limits as -3/2 and 3/2 - would this be correct?

Please help!

 

[daon2]

I have been over this problem a bunch of times and can't get it. can someone show me where I'm going wrong?

The problem asks, " find the volume of the solid... the wedge cut from the cylinder 4x² + y² = 9 by the planes z = 0 and z = y + 3

I started with the z = 0 base, and came up with z = 9 - 4x² - y² , which I set to be the f(x,y) in the middle of my integral statement. should I have incorporated the y + 3 here? where does that come into play? I have my integral set up to do type II (dA = dx dy) since I couldn't avoid a trig substitution otherwise, though maybe that was due to an earlier error. anyway so I have the limits of the x (inside) integral being x = (3/2 + y/16) for the upper limit and -(3/2 + y/16) for the lower limit, which I deduced from the ellipse on the xy plane (when z = 0).

I first integrated 9 - 4x² - y² with respect to x with these limits, and got:
-8/3 (3/2 + y/16)³ - 3y² - y³/8.

I then integrated this with respect to y, the y limits being -3 and 3 (again from the ellipse in the xy plane). this somehow ended up in me getting 1755/32 for the volume. this is not correct.

The first thing I tried integrating as my f(x,y) was y + 3. was this the right way to go?

Should I maintain the limits I had? One iteration had me using the y limits as sqrt(9-4x²) and -sqrt(9-4x²), and then the x limits as -3/2 and 3/2 - would this be correct?

Please help!

z is going from 0 to y+3, not to 9-4x^2-y^2. If this were the case, the top of the region would be an elliptic paraboloid, but its a plane.

x and y are filling out the ellipse 4x^2+y^2=9. Let D be the solid, R the region in the xy-plane.

$\displaystyle \displaystyle V = \iiint_D 1 dV = \iint_R \left(\int_0^{y+3} dz \right)dA = \int_{-3/2}^{3/2}\ \int_{-\sqrt{9-4x^2}}^{\sqrt{9-4x^2}} \ (y+3)\ dy \ dx = \cdots$