Calculus 3: Surface Area and Triple Intergrals

1. Set up, but do not evaluate, a double integral representing the surface area of the portion of the paraboloid z = 16 - x^2 - 4y^2 in the first octant.

2. Evaluate the triple integral. (from left to right, top numbers are 9, y/3, and sqrt(y^2 - 9x^2), bottom numbers are all zero) z dz dx dy

(both are from Calculus Early Transcendental Functions 5e) 14.5 #16 and 14.6 #4

3. Find the volume of the solid bounded by the surface z = 25 - x^2 - y^2 and the plane z = 0

Re: Calculus 3: Surface Area and Triple Intergrals

Quote:

Originally Posted by

**drumcorpsguy** 1. Set up, but do not evaluate, a double integral representing the surface area of the portion of the paraboloid z = 16 - x^2 - 4y^2 in the first octant.

2. Evaluate the triple integral. (from left to right, top numbers are 9, y/3, and sqrt(y^2 - 9x^2), bottom numbers are all zero) z dz dx dy

(both are from Calculus Early Transcendental Functions 5e) 14.5 #16 and 14.6 #4

3. Find the volume of the solid bounded by the surface z = 25 - x^2 - y^2 and the plane z = 0

2.

Re: Calculus 3: Surface Area and Triple Intergrals

2. Should have a square root on the third integral. I got 729/4, which seems correct.

Re: Calculus 3: Surface Area and Triple Intergrals

Re: Calculus 3: Surface Area and Triple Intergrals

Any idea on the other two. My answers below.

1. rom left to right, top numbers are 4 and sqrt(4 - (1/4)x^2), bottom numbers are all zero) sqrt(1 + 4x^2 + 64y^2) dy dx

3. 625(pi)/2. Used calculator, but would like to see the work if possible.