Finding parametric surface area

[Nat3]

I was able to get an answer to this homework problem, but I have no way of verifying that it is correct. I was hoping someone more experienced than me could look over my work and let me

know if I did the problem correctly.

Homework Statement

Find the surface area of the part of the plane $$2x+5y+z=10$$ that lies inside the cylinder $$x^2+y^2 = 9$$

Homework Equations

Parametric surface area:
$$\vec r (u,v) = <x(u,v), y(u, v), z(u, v)>$$

$$A(S) = \iint\limits_D \left |\vec r_u\times \vec r_v \right |dA$$

The Attempt at a Solution

Let x = x, y = y, z = 10 - 2x - 5y

Then $$\vec r(u, v) = <x, y, 10-2x-5y>$$, let u = x and v = y

$$\vec r_x(x, y) = <1, 0, -2>, \vec r_y(x, y) = <0, 1, -5>$$

$$\vec r_x \times \vec r_y = \left | \begin{array}{ccc} \vec i & \vec j & \vec k \\ 1 & 0 & -2 \\ 0 & 1 & -5 \end{array} \right| = <2, 5, 1>$$


$$\left | \vec r_x \times \vec r_y \right | = \sqrt {2^2 + 5^2 + 1} = \sqrt {30}$$

$$A(S) = \iint\limits_D \left |\vec r_u\times \vec r_v \right |dA = \iint\limits_D \sqrt{30} dA$$
Where D is the region bounded by $$x^2+y^2 = 9$$, or in polar, $$r^2 = 9$$

$$A(S) = \iint\limits_D \sqrt{30}dA = \int\limits_0^{2\pi} \int\limits_0^3 \sqrt{30}\ rdrd\theta = \sqrt{30} \int\limits_0^{2\pi} \left. \frac{1}{2} r^2 \right |_0^3 d<br /> <br /> \theta = \frac{\sqrt{30}}{2} \int\limits_0^{2\pi} \left. 9\ d\theta = \frac{9\sqrt{30}}{2}\theta \left. |_0^{2\pi} \right = 9\sqrt{30}$$

 

[LCKurtz]

Looks good except you dropped a pi at the very last step.

 

[HallsofIvy]

I was able to get an answer to this homework problem, but I have no way of verifying that it is correct. I was hoping someone more experienced than me could look over my work and let me

know if I did the problem correctly.

Homework Statement

Find the surface area of the part of the plane $$2x+5y+z=10$$ that lies inside the cylinder $$x^2+y^2 = 9$$

Homework Equations

Parametric surface area:
$$\vec r (u,v) = <x(u,v), y(u, v), z(u, v)>$$

$$A(S) = \iint\limits_D \left |\vec r_u\times \vec r_v \right |dA$$

The Attempt at a Solution

Let x = x, y = y, z = 10 - 2x - 5y

Then $$\vec r(u, v) = <x, y, 10-2x-5y>$$, let u = x and v = y

$$\vec r_x(x, y) = <1, 0, -2>, \vec r_y(x, y) = <0, 1, -5>$$

$$\vec r_x \times \vec r_y = \left | \begin{array}{ccc} \vec i & \vec j & \vec k \\ 1 & 0 & -2 \\ 0 & 1 & -5 \end{array} \right| = <2, 5, 1>$$


$$\left | \vec r_x \times \vec r_y \right | = \sqrt {2^2 + 5^2 + 1} = \sqrt {30}$$

$$A(S) = \iint\limits_D \left |\vec r_u\times \vec r_v \right |dA = \iint\limits_D \sqrt{30} dA$$
Where D is the region bounded by $$x^2+y^2 = 9$$, or in polar, $$r^2 = 9$$

$$A(S) = \iint\limits_D \sqrt{30}dA = \int\limits_0^{2\pi} \int\limits_0^3 \sqrt{30}\ rdrd\theta = \sqrt{30} \int\limits_0^{2\pi} \left. \frac{1}{2} r^2 \right |_0^3 d<br /> <br /> \theta = \frac{\sqrt{30}}{2} \int\limits_0^{2\pi} \left. 9\ d\theta = \frac{9\sqrt{30}}{2}\theta \left. |_0^{2\pi} \right = 9\sqrt{30}$$

It is isn't really necessary to integrate. $\int\int dA= A$, the area. And the area of a disk of radius 3 is $9\pi$
$$\int\int \sqrt{30}dA= 9\pi \sqrt{30}$$

 

[Nat3]

Thanks!