Rapid Recall
Consider the parametric surface $\vec r(u,v) = (u\cos v, u\sin v, u^2)$ for $u\in [1,4]$ and $v\in [0,2\pi]$. This surface is obtained by revolving the curve $z=y^2$ about the $z$-axis.
- Compute a normal vector to the surface (so $\vec n = \vec r_u\times \vec r_v$).
Solution
We have $\vec r_u = (\cos v, \sin v, 2u)$ and $\vec r_v = (-u\sin v, u\cos v,0)$. Their cross product gives $$\vec n = \vec r_u\times \vec r_v = (-2u^2\cos v,-2u^2\sin v,u\cos^2v + u \sin^2 v) = (-2u^2\cos v, -2u^2\sin v, u).$$
- Give an equation of the tangent plane to the surface at $(u,v)=(3,\pi/2)$.
Solution
To get an equation of a plane, we need a normal vector $\vec n = (A,B,C)$ and a point $P=(a,b,c)$. An equation of the plane is then given by $$A(x-a)+B(y-b)+C(z-c)=0.$$ Recall this means $\vec n$ and $\vec {PQ}$ are orthogonal, where $Q=(x,y,z)$ represents any point in the plane.
A point is $$\vec r(3,\pi/2) = (3\cos(\pi/2), 3\sin(\pi/2), 3^2) = (0, 3, 9).$$ A normal vector is $$\vec n(3,\pi/2) = (-2(3)^2\cos (\pi/2), -2(3)^2\sin (\pi/2), 3) = (0, -18, 3).$$ This means an equation of the tangent plane is $$0(x-0)-18(y-3)+3(z-9)=0.$$
- Set up an integral to compute the surface area of the surface above.
Solution
To get surface area $\sigma$, we need to add up little bits of surface area $d\sigma$. To find a small bit of surface area, we use the partial derivatives from above to form tiny parallelograms on the surface. The area of these parallelograms is $$d\sigma = |\vec r_u\times \vec r_v|dudv.$$ So we just need to find the length of $\vec n$ from above. We obtain $$d\sigma = |\vec r_u\times \vec r_v|dudv = \sqrt{4u^4\cos^2 v+4u^4\sin^2v+u^2}dudv =\sqrt{4u^4+u^2}dudv = u\sqrt{4u^2+1}dudv.$$ The surface area is then $$\sigma = \iint_S d\sigma = \int_0^{2\pi}\int_1^4 u\sqrt{4u^2+1}dudv.$$
- Set up an integral to compute $\bar z$ for the surface above.
Solution
We've done most of the work above. We just need to apply the average value formula. Note that we get $z = u^2$ from the original parametrization. This gives $$\bar z = \frac{\iint_S z d\sigma}{\iint_S d\sigma} = \frac{\int_0^{2\pi}\int_1^4 (u^2)u\sqrt{4u^2+1}dudv}{\int_0^{2\pi}\int_1^4 u\sqrt{4u^2+1}dudv} .$$
Group problems
- Consider the parametric surface $\vec r(u,v) = (u^2\cos v, u, u^2\sin v)$ for $u\in [1,4]$ and $v\in [0,2\pi]$. This surface is obtained by revolving the curve $z=y^2$ about the $y$-axis.
- Compute a normal vector to the surface (so $\vec n = \vec r_u\times \vec r_v$).
- Give an equation of the tangent plane to the surface at $(u,v)=(3,\pi/2)$.
- Set up an integral to compute the surface area of the surface.
- Consider the surface parametrized by $\vec r(u,v) = (u, v, u^2+v^2)$ for $-3\leq u\leq 3$ and $0\leq v\leq 3$.
- Compute $d\sigma = \left |\dfrac{\partial \vec r}{\partial u}\times\dfrac{\partial \vec r}{\partial u}\right|dudv$.
- Set up an integral formula to compute $\bar z$ for this surface.
- Draw each curve or surface given below. If you want practice setting up line or surface integrals, set up an integral to compute the length (if a curve) or surface area (if a surface). What helps you realize it is a curve or a surface?
- $\vec r(u,v) = (4\cos u,v, 3\sin u)$ for $0\leq u\leq \pi$ and $0\leq v\leq 7$.
- $\vec r(t) = (3\cos t,3\sin t,t)$ for $0\leq t\leq 6\pi$.
- $\vec r(u,v) = (u\cos v,u\sin v,v)$ for $0\leq v\leq 6\pi$ and $2\leq u\leq 4$.
- $\vec r(t) = (0,t,9-t^2)$ for $0\leq t\leq 3$.
- $\vec r(x,y) = (x,y,9-x^2-y^2)$ for for $0\leq x \leq 3$ and $-3\leq y\leq 3$.
- $\vec r(u,v) = (u\cos v,u\sin v,9-u^2)$ for $0\leq u\leq 3$ and $0\leq v\leq 2\pi$.
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