191. y = x y = x from x = 2 x = 2 to x = 6 x = 6. &= -55 \int_0^{2\pi} du \4pt] There is more to this sketch than the actual surface itself. To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. To approximate the mass flux across $$S$$, form the sum, \[\sum_{i=1}m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. Some surfaces, such as a Mbius strip, cannot be oriented. Since $$S_{ij}$$ is small, the dot product $$\rho v \cdot N$$ changes very little as we vary across $$S_{ij}$$ and therefore $$\rho \vecs v \cdot \vecs N$$ can be taken as approximately constant across $$S_{ij}$$. Analogously, we would like a notion of regularity (or smoothness) for surfaces so that a surface parameterization really does trace out a surface. Before calculating any integrals, note that the gradient of the temperature is $$\vecs \nabla T = \langle 2xz, \, 2yz, \, x^2 + y^2 \rangle$$. What Is a Surface Area Calculator in Calculus? Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. To embed a widget in your blog's sidebar, install the Wolfram|Alpha Widget Sidebar Plugin, and copy and paste the Widget ID below into the "id" field: We appreciate your interest in Wolfram|Alpha and will be in touch soon. , for which the given function is differentiated. Therefore, \[\begin{align*} \iint_{S_1} z^2 \,dS &= \int_0^{\sqrt{3}} \int_0^{2\pi} f(r(u,v))||t_u \times t_v|| \, dv \, du \\ Find the surface area of the surface with parameterization $$\vecs r(u,v) = \langle u + v, \, u^2, \, 2v \rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 2$$. Therefore, a parameterization of this cone is, \[\vecs s(u,v) = \langle kv \, \cos u, \, kv \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h. \nonumber. For example, let's say you want to calculate the magnitude of the electric flux through a closed surface around a 10 n C 10\ \mathrm{nC} 10 nC electric charge. Since every curve has a forward and backward direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. and $$||\vecs t_u \times \vecs t_v || = \sqrt{\cos^2 u + \sin^2 u} = 1$$. First, a parser analyzes the mathematical function. Let $$S$$ be hemisphere $$x^2 + y^2 + z^2 = 9$$ with $$z \leq 0$$ such that $$S$$ is oriented outward. \nonumber \], For grid curve $$\vecs r(u, v_j)$$, the tangent vector at $$P_{ij}$$ is, $\vecs t_u (P_{ij}) = \vecs r_u (u_i,v_j) = \langle x_u (u_i,v_j), \, y_u(u_i,v_j), \, z_u (u_i,v_j) \rangle. n d . Exercise12.1.8 For both parts of this exercise, the computations involved were actually done in previous problems. Calculus: Fundamental Theorem of Calculus &= \rho^2 \sin^2 \phi (\cos^2 \theta + \sin^2 \theta) \\[4pt] Multiple Integrals Calculator - Symbolab Multiple Integrals Calculator Solve multiple integrals step-by-step full pad Examples Related Symbolab blog posts Advanced Math Solutions - Integral Calculator, trigonometric substitution In the previous posts we covered substitution, but standard substitution is not always enough. If $$v = 0$$ or $$v = \pi$$, then the only choices for $$u$$ that make the $$\mathbf{\hat{j}}$$ component zero are $$u = 0$$ or $$u = \pi$$. The surface integral of the vector field over the oriented surface (or the flux of the vector field across First we calculate the partial derivatives:. That's why showing the steps of calculation is very challenging for integrals. For any point $$(x,y,z)$$ on $$S$$, we can identify two unit normal vectors $$\vecs N$$ and $$-\vecs N$$. First, we are using pretty much the same surface (the integrand is different however) as the previous example. Surface integrals of scalar fields. New Resources. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero. Therefore, the surface is the elliptic paraboloid $$x^2 + y^2 = z$$ (Figure $$\PageIndex{3}$$). In the next block, the lower limit of the given function is entered. Find the ux of F = zi +xj +yk outward through the portion of the cylinder This time, the function gets transformed into a form that can be understood by the computer algebra system Maxima. This surface has parameterization $$\vecs r(x, \theta) = \langle x, \, x^2 \cos \theta, \, x^2 \sin \theta \rangle, \, 0 \leq x \leq b, \, 0 \leq x < 2\pi.$$. Why do you add a function to the integral of surface integrals? The tangent vectors are $$\vecs t_u = \langle 1,-1,1\rangle$$ and $$\vecs t_v = \langle 0,2v,1\rangle$$. The entire surface is created by making all possible choices of $$u$$ and $$v$$ over the parameter domain. There are two moments, denoted by M x M x and M y M y. This is called the positive orientation of the closed surface (Figure $$\PageIndex{18}$$). is a dot product and is a unit normal vector. Last, lets consider the cylindrical side of the object. Therefore, to calculate, \[\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber$. Describe the surface with parameterization, $\vecs{r} (u,v) = \langle 2 \, \cos u, \, 2 \, \sin u, \, v \rangle, \, 0 \leq u \leq 2\pi, \, -\infty < v < \infty \nonumber$. Notice that if we change the parameter domain, we could get a different surface. Choose point $$P_{ij}$$ in each piece $$S_{ij}$$ evaluate $$P_{ij}$$ at $$f$$, and multiply by area $$S_{ij}$$ to form the Riemann sum, $\sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \, \Delta S_{ij}. \nonumber$. &=80 \int_0^{2\pi} 45 \, d\theta \\ We can drop the absolute value bars in the sine because sine is positive in the range of $$\varphi$$ that we are working with. In particular, surface integrals allow us to generalize Greens theorem to higher dimensions, and they appear in some important theorems we discuss in later sections. The image of this parameterization is simply point $$(1,2)$$, which is not a curve. Like so many things in multivariable calculus, while the theory behind surface integrals is beautiful, actually computing one can be painfully labor intensive. The practice problem generator allows you to generate as many random exercises as you want. Find the flux of F = y z j ^ + z 2 k ^ outward through the surface S cut from the cylinder y 2 + z 2 = 1, z 0, by the planes x = 0 and x = 1. A cast-iron solid ball is given by inequality $$x^2 + y^2 + z^2 \leq 1$$. Again, notice the similarities between this definition and the definition of a scalar line integral. Furthermore, assume that $$S$$ is traced out only once as $$(u,v)$$ varies over $$D$$. A parameterized surface is given by a description of the form, $\vecs{r}(u,v) = \langle x (u,v), \, y(u,v), \, z(u,v)\rangle. Therefore, the tangent of $$\phi$$ is $$\sqrt{3}$$, which implies that $$\phi$$ is $$\pi / 6$$. \nonumber$. Surface Integral of a Vector Field. Double Integral calculator with Steps & Solver It can be also used to calculate the volume under the surface. The Surface Area Calculator uses a formula using the upper and lower limits of the function for the axis along which the arc revolves. To be precise, consider the grid lines that go through point $$(u_i, v_j)$$. we can always use this form for these kinds of surfaces as well. Find step by step results, graphs & plot using multiple integrals, Step 1: Enter the function and the limits in the input field Step 2: Now click the button Calculate to get the value Step 3: Finally, the, For a scalar function f over a surface parameterized by u and v, the surface integral is given by Phi = int_Sfda (1) = int_Sf(u,v)|T_uxT_v|dudv. Suppose that $$i$$ ranges from $$1$$ to $$m$$ and $$j$$ ranges from $$1$$ to $$n$$ so that $$D$$ is subdivided into $$mn$$ rectangles. Therefore, we calculate three separate integrals, one for each smooth piece of $$S$$. The partial derivatives in the formulas are calculated in the following way: Hold $$u$$ constant and see what kind of curves result. Therefore, a point on the cone at height $$u$$ has coordinates $$(u \, \cos v, \, u \, \sin v, \, u)$$ for angle $$v$$. For example, consider curve parameterization $$\vecs r(t) = \langle 1,2\rangle, \, 0 \leq t \leq 5$$. \nonumber \]. . They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. Step #2: Select the variable as X or Y. The boundary curve, C , is oriented clockwise when looking along the positive y-axis. If we think of $$\vecs r$$ as a mapping from the $$uv$$-plane to $$\mathbb{R}^3$$, the grid curves are the image of the grid lines under $$\vecs r$$. \nonumber \]. Integral $$\displaystyle \iint_S \vecs F \cdot \vecs N\, dS$$ is called the flux of $$\vecs{F}$$ across $$S$$, just as integral $$\displaystyle \int_C \vecs F \cdot \vecs N\,dS$$ is the flux of $$\vecs F$$ across curve $$C$$. This page titled 16.6: Surface Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin Jed Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Calculate the lateral surface area (the area of the side, not including the base) of the right circular cone with height h and radius r. Before calculating the surface area of this cone using Equation \ref{equation1}, we need a parameterization. \nonumber \]. The program that does this has been developed over several years and is written in Maxima's own programming language. However, before we can integrate over a surface, we need to consider the surface itself. Surface integrals are a generalization of line integrals. Investigate the cross product $$\vecs r_u \times \vecs r_v$$. Hold $$u$$ and $$v$$ constant, and see what kind of curves result. The reason for this is that the circular base is included as part of the cone, and therefore the area of the base $$\pi r^2$$ is added to the lateral surface area $$\pi r \sqrt{h^2 + r^2}$$ that we found. The tangent vectors are $$\vecs t_x = \langle 1,0,1 \rangle$$ and $$\vecs t_y = \langle 1,0,2 \rangle$$. Then, $$\vecs t_x = \langle 1,0,f_x \rangle$$ and $$\vecs t_y = \langle 0,1,f_y \rangle$$, and therefore the cross product $$\vecs t_x \times \vecs t_y$$ (which is normal to the surface at any point on the surface) is $$\langle -f_x, \, -f_y, \, 1 \rangle$$Since the $$z$$-component of this vector is one, the corresponding unit normal vector points upward, and the upward side of the surface is chosen to be the positive side. We discuss how Surface integral of vector field calculator can help students learn Algebra in this blog post. Solution : Since we are given a line integral and told to use Stokes' theorem, we need to compute a surface integral. Send feedback | Visit Wolfram|Alpha. Compute the net mass outflow through the cube formed by the planes x=0, x=1, y=0, y=1, z=0, z=1. \end{align*}\]. For grid curve $$\vecs r(u_i,v)$$, the tangent vector at $$P_{ij}$$ is, $\vecs t_v (P_{ij}) = \vecs r_v (u_i,v_j) = \langle x_v (u_i,v_j), \, y_v(u_i,v_j), \, z_v (u_i,v_j) \rangle. $$\vecs t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle$$ and $$\vecs t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle$$, and $$\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0, \, 0, -v \rangle$$. Step #4: Fill in the lower bound value. Calculate surface integral \[\iint_S f(x,y,z)\,dS, \nonumber$ where $$f(x,y,z) = z^2$$ and $$S$$ is the surface that consists of the piece of sphere $$x^2 + y^2 + z^2 = 4$$ that lies on or above plane $$z = 1$$ and the disk that is enclosed by intersection plane $$z = 1$$ and the given sphere (Figure $$\PageIndex{16}$$). By Equation, the heat flow across $$S_1$$ is, \begin{align*}\iint_{S_2} -k \vecs \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_0^1 \vecs \nabla T(u,v) \cdot\, (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt] How can we calculate the amount of a vector field that flows through common surfaces, such as the . Therefore, \[\vecs t_u \times \vecs t_v = \langle -1 -2v, -1, 2v\rangle. The parameterization of the cylinder and $$\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|$$ is. Integrate the work along the section of the path from t = a to t = b. If $$u = v = 0$$, then $$\vecs r(0,0) = \langle 1,0,0 \rangle$$, so point (1, 0, 0) is on $$S$$. Since the disk is formed where plane $$z = 1$$ intersects sphere $$x^2 + y^2 + z^2 = 4$$, we can substitute $$z = 1$$ into equation $$x^2 + y^2 + z^2 = 4$$: \[x^2 + y^2 + 1 = 4 \Rightarrow x^2 + y^2 = 3. \end{align*}. Choose point $$P_{ij}$$ in each piece $$S_{ij}$$. This helps me sooo much, im in seventh grade and this helps A LOT, i was able to pass and ixl in 3 minutes, and it was a word problems one. A surface integral of a vector field. To get an orientation of the surface, we compute the unit normal vector, In this case, $$\vecs t_u \times \vecs t_v = \langle r \, \cos u, \, r \, \sin u, \, 0 \rangle$$ and therefore, $||\vecs t_u \times \vecs t_v|| = \sqrt{r^2 \cos^2 u + r^2 \sin^2 u} = r. \nonumber$, $\vecs N(u,v) = \dfrac{\langle r \, \cos u, \, r \, \sin u, \, 0 \rangle }{r} = \langle \cos u, \, \sin u, \, 0 \rangle. where &= \iint_D (\vecs F(\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v))\,dA. &= - 55 \int_0^{2\pi} \int_1^4 \langle 2v \, \cos u, \, 2v \, \sin u, \, \cos^2 u + \sin^2 u \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\, du \\[4pt] In the case of the y-axis, it is c. Against the block titled to, the upper limit of the given function is entered. If you don't specify the bounds, only the antiderivative will be computed. However, if I have a numerical integral then I can just make . Since $$S$$ is given by the function $$f(x,y) = 1 + x + 2y$$, a parameterization of $$S$$ is $$\vecs r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2$$. So, we want to find the center of mass of the region below. Hence, a parameterization of the cone is $$\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle$$. Note how the equation for a surface integral is similar to the equation for the line integral of a vector field C F d s = a b F ( c ( t)) c ( t) d t. For line integrals, we integrate the component of the vector field in the tangent direction given by c ( t). For each point $$\vecs r(a,b)$$ on the surface, vectors $$\vecs t_u$$ and $$\vecs t_v$$ lie in the tangent plane at that point. It's just a matter of smooshing the two intuitions together. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral, \[\iint_S \vecs v \cdot \vecs N \, dS, \nonumber$. uses a formula using the upper and lower limits of the function for the axis along which the arc revolves. 6.6.1 Find the parametric representations of a cylinder, a cone, and a sphere. Therefore, the choice of unit normal vector, $\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \nonumber$. Now consider the vectors that are tangent to these grid curves. Make sure that it shows exactly what you want. Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. So, for our example we will have. In particular, they are used for calculations of. After putting the value of the function y and the lower and upper limits in the required blocks, the result appears as follows: $S = \int_{1}^{2} 2 \pi x^2 \sqrt{1+ (\dfrac{d(x^2)}{dx})^2}\, dx$, $S = \dfrac{1}{32} pi (-18\sqrt{5} + 132\sqrt{17} + sinh^{-1}(2) sinh^{-1}(4))$. Skip the "f(x) =" part and the differential "dx"! Each choice of $$u$$ and $$v$$ in the parameter domain gives a point on the surface, just as each choice of a parameter $$t$$ gives a point on a parameterized curve. Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure $$\PageIndex{7}$$). &= 7200\pi.\end{align*} \nonumber \]. Integrations is used in various fields such as engineering to determine the shape and size of strcutures. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. &= 32 \pi \left[ \dfrac{1}{3} - \dfrac{\sqrt{3}}{8} \right] = \dfrac{32\pi}{3} - 4\sqrt{3}. We see that $$S_2$$ is a circle of radius 1 centered at point $$(0,0,4)$$, sitting in plane $$z = 4$$. Notice that we plugged in the equation of the plane for the x in the integrand. Describe the surface integral of a scalar-valued function over a parametric surface. Find a parameterization r ( t) for the curve C for interval t. Find the tangent vector. If you have any questions or ideas for improvements to the Integral Calculator, don't hesitate to write me an e-mail. Calculate the area of a surface of revolution step by step The calculations and the answer for the integral can be seen here. By Example, we know that $$\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle$$. Schematic representation of a surface integral The surface integral is calculated by taking the integral of the dot product of the vector field with The calculator lacks the mathematical intuition that is very useful for finding an antiderivative, but on the other hand it can try a large number of possibilities within a short amount of time. &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 \, \sin^3 \phi + 27 \, \cos^2 \phi \, \sin \phi \, d\phi \, d\theta \\ The rate of flow, measured in mass per unit time per unit area, is $$\rho \vecs N$$. The Divergence Theorem can be also written in coordinate form as. Parameterize the surface and use the fact that the surface is the graph of a function. To use Equation \ref{scalar surface integrals} to calculate the surface integral, we first find vectors $$\vecs t_u$$ and $$\vecs t_v$$. &= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4b^2 + 1} (8b^3 + b) \, \sinh^{-1} (2b) \right)\right]. The surface is a portion of the sphere of radius 2 centered at the origin, in fact exactly one-eighth of the sphere. $\vecs r(\phi, \theta) = \langle 3 \, \cos \theta \, \sin \phi, \, 3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi/2. Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more. It helps you practice by showing you the full working (step by step integration). Is the surface parameterization $$\vecs r(u,v) = \langle u^{2v}, v + 1, \, \sin u \rangle, \, 0 \leq u \leq 2, \, 0 \leq v \leq 3$$ smooth? All common integration techniques and even special functions are supported. Substitute the parameterization into F . Loading please wait!This will take a few seconds. \label{equation 5}$, $\iint_S \vecs F \cdot \vecs N\,dS, \nonumber$, where $$\vecs{F} = \langle -y,x,0\rangle$$ and $$S$$ is the surface with parameterization, $\vecs r(u,v) = \langle u,v^2 - u, \, u + v\rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 4. Then, $$S$$ can be parameterized with parameters $$x$$ and $$\theta$$ by, \[\vecs r(x, \theta) = \langle x, f(x) \, \cos \theta, \, f(x) \sin \theta \rangle, \, a \leq x \leq b, \, 0 \leq x \leq 2\pi. the parameter domain of the parameterization is the set of points in the $$uv$$-plane that can be substituted into $$\vecs r$$. The integration by parts calculator is simple and easy to use. It's like with triple integrals, how you use them for volume computations a lot, but in their full glory they can associate any function with a 3-d region, not just the function f(x,y,z)=1, which is how the volume computation ends up going. Surface Area Calculator Author: Ravinder Kumar Topic: Area, Surface The present GeoGebra applet shows surface area generated by rotating an arc. Since the parameter domain is all of $$\mathbb{R}^2$$, we can choose any value for u and v and plot the corresponding point. In "Options", you can set the variable of integration and the integration bounds. Computing a surface integral is almost identical to computing surface area using a double integral, except that you stick a function inside the integral. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \sqrt{\sin^2\phi + \cos^2\phi} \, d\phi \\ The surface integral will have a $$dS$$ while the standard double integral will have a $$dA$$. In this video we come up formulas for surface integrals, which are when we accumulate the values of a scalar function over a surface. &= 32\pi \left[- \dfrac{\cos^3 \phi}{3} \right]_0^{\pi/6} \\ A surface parameterization $$\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$$ is smooth if vector $$\vecs r_u \times \vecs r_v$$ is not zero for any choice of $$u$$ and $$v$$ in the parameter domain. We have seen that a line integral is an integral over a path in a plane or in space. This surface has parameterization $$\vecs r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4$$. These grid lines correspond to a set of grid curves on surface $$S$$ that is parameterized by $$\vecs r(u,v)$$. This can also be written compactly in vector form as (2) If the region is on the left when traveling around , then area of can be computed using the elegant formula (3) To calculate the mass flux across $$S$$, chop $$S$$ into small pieces $$S_{ij}$$. However, since we are on the cylinder we know what $$y$$ is from the parameterization so we will also need to plug that in. Surface Integral -- from Wolfram MathWorld Calculus and Analysis Differential Geometry Differential Geometry of Surfaces Algebra Vector Algebra Calculus and Analysis Integrals Definite Integrals Surface Integral For a scalar function over a surface parameterized by and , the surface integral is given by (1) (2) When the "Go!" &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. Lets now generalize the notions of smoothness and regularity to a parametric surface. Step 2: Compute the area of each piece. What if you are considering the surface of a curved airplane wing with variable density, and you want to find its total mass? surface integral Natural Language Math Input Use Math Input Mode to directly enter textbook math notation. First, lets look at the surface integral of a scalar-valued function. &= \rho^2 \, \sin^2 \phi \\[4pt] The following theorem provides an easier way in the case when  is a closed surface, that is, when  encloses a bounded solid in $$\mathbb{R}^ 3$$. Let C be the closed curve illustrated below. Therefore, the area of the parallelogram used to approximate the area of $$S_{ij}$$ is, \[\Delta S_{ij} \approx ||(\Delta u \vecs t_u (P_{ij})) \times (\Delta v \vecs t_v (P_{ij})) || = ||\vecs t_u (P_{ij}) \times \vecs t_v (P_{ij}) || \Delta u \,\Delta v. \nonumber$. We rewrite the equation of the plane in the form Find the partial derivatives: Applying the formula we can express the surface integral in terms of the double integral: The region of integration is the triangle shown in Figure Figure 2. and Now it is time for a surface integral example: For more on surface area check my online book "Flipped Classroom Calculus of Single Variable" https://versal.com/learn/vh45au/ Step #5: Click on "CALCULATE" button. Let's take a closer look at each form . The second step is to define the surface area of a parametric surface. What about surface integrals over a vector field? Now we need $${\vec r_z} \times {\vec r_\theta }$$. Find the area of the surface of revolution obtained by rotating $$y = x^2, \, 0 \leq x \leq b$$ about the x-axis (Figure $$\PageIndex{14}$$). why is michael severe leaving 1620 the zone, uconn staff directory,