Calculus 3: Surface Area and Triple Intergrals

• Dec 12th 2011, 08:46 AM
drumcorpsguy
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
• Dec 12th 2011, 04:34 PM
Prove It
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.
\displaystyle \displaystyle \begin{align*} \int_0^9{\int_0^{\frac{y}{3}}{\int_0^{y^2-9x^2}{z\,dz}\,dx}\,dy} &= \int_0^9{\int_0^{\frac{y}{3}}{\left[\frac{z^2}{2}\right]_0^{y^2 - 9x^2}\,dx}\,dy} \\ &= \int_0^9{\int_0^{\frac{y}{3}}{\frac{\left(y^2 - 9x^2\right)^2}{2} - \frac{0^2}{2}\,dx}\,dy} \\ &= \frac{1}{2}\int_0^9{\int_0^{\frac{y}{3}}{y^4 - 18x^2y^2 + 81x^4\,dx}\,dy} \\ &= \frac{1}{2}\int_0^9{\left[x\,y^4 - 6x^3y^2 + \frac{81x^5}{5}\right]_0^{\frac{y}{3}}\,dy} \\ &= \frac{1}{2}\int_0^9{\left[\left(\frac{y}{3}\right)y^4 - 6\left(\frac{y}{3}\right)^3y^2 + \frac{81\left(\frac{y}{3}\right)^5}{5}\right] - \left[(0)y^4 - 6(0)^3y^2 + \frac{81(0)^5}{5}\right]\,dy} \\ &= \frac{1}{2}\int_0^9{\frac{y^5}{3} - \frac{2y^5}{9} + \frac{y^5}{15} \,dy} \\ &= \frac{1}{90}\int_0^9{15y^5 - 10y^5 + 3y^5\,dy} \\ &= \frac{1}{90}\int_0^9{8y^5\,dy} \\ &= \frac{4}{45}\int_0^9{y^5\,dy} \\ &= \frac{4}{45}\left[\frac{y^6}{6}\right]_0^9\end{align*}

\displaystyle \displaystyle \begin{align*} &= \frac{4}{45}\left(\frac{9^6}{6} - \frac{0^6}{6}\right) \\ &= \frac{4}{45}\left(\frac{531\,441}{6}\right) \\ &= \frac{39\,366}{5} \end{align*}
• Dec 12th 2011, 05:47 PM
drumcorpsguy
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.
• Dec 12th 2011, 05:49 PM
Prove It
Re: Calculus 3: Surface Area and Triple Intergrals
You're welcome...
• Dec 12th 2011, 06:02 PM
drumcorpsguy
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.