and , In the case of antiderivatives, the entire procedure is repeated with each function's derivative, since antiderivatives are allowed to differ by a constant. Use parentheses! If , Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. If S is a cylinder given by equation \(x^2 + y^2 = R^2\), then a parameterization of \(S\) is \(\vecs r(u,v) = \langle R \, \cos u, \, R \, \sin u, \, v \rangle, \, 0 \leq u \leq 2 \pi, \, -\infty < v < \infty.\). This calculator consists of input boxes in which the values of the functions and the axis along which the revolution occurs are entered. You can use this calculator by first entering the given function and then the variables you want to differentiate against. Since \(S\) is given by the function \(f(x,y) = 1 + x + 2y\), a parameterization of \(S\) is \(\vecs r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2\). the parameter domain of the parameterization is the set of points in the \(uv\)-plane that can be substituted into \(\vecs r\). For example, the graph of \(f(x,y) = x^2 y\) can be parameterized by \(\vecs r(x,y) = \langle x,y,x^2y \rangle\), where the parameters \(x\) and \(y\) vary over the domain of \(f\). Direct link to benvessely's post Wow what you're crazy sma. ; 6.6.2 Describe the surface integral of a scalar-valued function over a parametric surface. Step 1: Chop up the surface into little pieces. For a curve, this condition ensures that the image of \(\vecs r\) really is a curve, and not just a point. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. This time, the function gets transformed into a form that can be understood by the computer algebra system Maxima. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces. First, lets look at the surface integral of a scalar-valued function. Conversely, each point on the cylinder is contained in some circle \(\langle \cos u, \, \sin u, \, k \rangle \) for some \(k\), and therefore each point on the cylinder is contained in the parameterized surface (Figure \(\PageIndex{2}\)). &= 7200\pi.\end{align*} \nonumber \]. Similarly, the average value of a function of two variables over the rectangular The vendor states an area of 200 sq cm. In this section we introduce the idea of a surface integral. Note that all four surfaces of this solid are included in S S. Solution. The surface element contains information on both the area and the orientation of the surface. To calculate the surface integral, we first need a parameterization of the cylinder. Therefore, as \(u\) increases, the radius of the resulting circle increases. Parameterize the surface and use the fact that the surface is the graph of a function. This is not an issue though, because Equation \ref{scalar surface integrals} does not place any restrictions on the shape of the parameter domain. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners. First, we are using pretty much the same surface (the integrand is different however) as the previous example. We now have a parameterization of \(S_2\): \(\vecs r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi / 3.\), The tangent vectors are \(\vecs t_{\phi} = \langle 2 \, \cos \theta \, \cos \phi, \, 2 \, \sin \theta \,\cos \phi, \, -2 \, \sin \phi \rangle\) and \(\vecs t_{\theta} = \langle - 2 \sin \theta \sin \phi, \, u\cos \theta \sin \phi, \, 0 \rangle\), and thus, \[\begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 2 \cos \theta \cos \phi & 2 \sin \theta \cos \phi & -2\sin \phi \\ -2\sin \theta\sin\phi & 2\cos \theta \sin\phi & 0 \end{vmatrix} \\[4 pt] surface integral Natural Language Math Input Use Math Input Mode to directly enter textbook math notation. The surface integral of \(\vecs{F}\) over \(S\) is, \[\iint_S \vecs{F} \cdot \vecs{S} = \iint_S \vecs{F} \cdot \vecs{N} \,dS. Letting the vector field \(\rho \vecs{v}\) be an arbitrary vector field \(\vecs{F}\) leads to the following definition. We now show how to calculate the ux integral, beginning with two surfaces where n and dS are easy to calculate the cylinder and the sphere. This is the two-dimensional analog of line integrals. Varying point \(P_{ij}\) over all pieces \(S_{ij}\) and the previous approximation leads to the following definition of surface area of a parametric surface (Figure \(\PageIndex{11}\)). Posted 5 years ago. Double integral calculator with steps help you evaluate integrals online. to denote the surface integral, as in (3). Thus, a surface integral is similar to a line integral but in one higher dimension. &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54\, \sin \phi - 27 \, \cos^2 \phi \, \sin \phi \, d\phi \,d\theta \\ Take the dot product of the force and the tangent vector. 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. Surface integrals of vector fields. For a scalar function over a surface parameterized by and , the surface integral is given by. We assume here and throughout that the surface parameterization \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle\) is continuously differentiablemeaning, each component function has continuous partial derivatives. &= -55 \int_0^{2\pi} du \\[4pt] It consists of more than 17000 lines of code. \end{align*}\], Therefore, to compute a surface integral over a vector field we can use the equation, \[\iint_S \vecs F \cdot \vecs N\, dS = \iint_D (\vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v)) \,dA. If parameterization \(\vec{r}\) is regular, then the image of \(\vec{r}\) is a two-dimensional object, as a surface should be. All you need to do is to follow below steps: Step #1: Fill in the integral equation you want to solve. &= 5 \left[\dfrac{(1+4u^2)^{3/2}}{3} \right]_0^2 \\ Notice that this parameter domain \(D\) is a triangle, and therefore the parameter domain is not rectangular. Then, the unit normal vector is given by \(\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||}\) and, from Equation \ref{surfaceI}, we have, \[\begin{align*} \int_C \vecs F \cdot \vecs N\, dS &= \iint_S \vecs F \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \,dS \\[4pt] Our calculator allows you to check your solutions to calculus exercises. Example 1. &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos \phi \, \sin \phi \rangle. Without loss of generality, we assume that \(P_{ij}\) is located at the corner of two grid curves, as in Figure \(\PageIndex{9}\). Okay, since we are looking for the portion of the plane that lies in front of the \(yz\)-plane we are going to need to write the equation of the surface in the form \(x = g\left( {y,z} \right)\). Calculate surface integral \[\iint_S \vecs F \cdot \vecs N \, dS, \nonumber \] where \(\vecs F = \langle 0, -z, y \rangle\) and \(S\) is the portion of the unit sphere in the first octant with outward orientation. At the center point of the long dimension, it appears that the area below the line is about twice that above. Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure \(\PageIndex{7}\)). There are two moments, denoted by M x M x and M y M y. Math Assignments. When you're done entering your function, click "Go! Notice that \(S\) is not smooth but is piecewise smooth; \(S\) can be written as the union of its base \(S_1\) and its spherical top \(S_2\), and both \(S_1\) and \(S_2\) are smooth. To get an idea of the shape of the surface, we first plot some points. Very useful and convenient. These grid lines correspond to a set of grid curves on surface \(S\) that is parameterized by \(\vecs r(u,v)\). \end{align*}\]. We need to be careful here. \[S = \int_{0}^{4} 2 \pi y^{\dfrac1{4}} \sqrt{1+ (\dfrac{d(y^{\dfrac1{4}})}{dy})^2}\, dy \]. \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. By Example, we know that \(\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle\). Step #2: Select the variable as X or Y. Solution. For a vector function over a surface, the surface The integration by parts calculator is simple and easy to use. Our integral solver also displays anti-derivative calculations to users who might be interested in the mathematical concept and steps involved in integration. Integrals can be a little daunting for students, but they are essential to calculus and understanding more advanced mathematics. This idea of adding up values over a continuous two-dimensional region can be useful for curved surfaces as well. Having an integrand allows for more possibilities with what the integral can do for you. Suppose that \(v\) is a constant \(K\). The \(\mathbf{\hat{k}}\) component of this vector is zero only if \(v = 0\) or \(v = \pi\). It's just a matter of smooshing the two intuitions together. To be precise, consider the grid lines that go through point \((u_i, v_j)\). &= 2\pi \sqrt{3}. The parameterization of full sphere \(x^2 + y^2 + z^2 = 4\) is, \[\vecs r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi. For example, this involves writing trigonometric/hyperbolic functions in their exponential forms. &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos^2 u, \, 2v \, \sin u, \, 1 \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\,\, du \\[4pt] 0y4 and the rotation are along the y-axis. Put the value of the function and the lower and upper limits in the required blocks on the calculator t, Surface Area Calculator Calculus + Online Solver With Free Steps. Interactive graphs/plots help visualize and better understand the functions. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \sqrt{\sin^2\phi + \cos^2\phi} \, d\phi \\ I have been tasked with solving surface integral of ${\bf V} = x^2{\bf e_x}+ y^2{\bf e_y}+ z^2 {\bf e_z}$ on the surface of a cube bounding the region $0\le x,y,z \le 1$. Therefore, the mass flux is, \[\iint_s \rho \vecs v \cdot \vecs N \, dS = \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. \nonumber \]. Thank you! It transforms it into a form that is better understandable by a computer, namely a tree (see figure below). The program that does this has been developed over several years and is written in Maxima's own programming language. C F d s. using Stokes' Theorem. Just as with vector line integrals, surface integral \(\displaystyle \iint_S \vecs F \cdot \vecs N\, dS\) is easier to compute after surface \(S\) has been parameterized. \end{align*}\], Calculate \[\iint_S (x^2 - z) \,dS, \nonumber \] where \(S\) is the surface with parameterization \(\vecs r(u,v) = \langle v, \, u^2 + v^2, \, 1 \rangle, \, 0 \leq u \leq 2, \, 0 \leq v \leq 3.\). This is an easy surface integral to calculate using the Divergence Theorem: $$ \iiint_E {\rm div} (F)\ dV = \iint_ {S=\partial E} \vec {F}\cdot d {\bf S}$$ However, to confirm the divergence theorem by the direct calculation of the surface integral, how should the bounds on the double integral for a unit ball be chosen? The surface integral will have a dS d S while the standard double integral will have a dA d A. The image of this parameterization is simply point \((1,2)\), which is not a curve. 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. https://mathworld.wolfram.com/SurfaceIntegral.html. 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. &= \sqrt{6} \int_0^4 \dfrac{22x^2}{3} + 2x^3 \,dx \\[4pt] Therefore, we calculate three separate integrals, one for each smooth piece of \(S\). \nonumber \]. We will see one of these formulas in the examples and well leave the other to you to write down. That is: To make the work easier I use the divergence theorem, to replace the surface integral with a . The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Use the standard parameterization of a cylinder and follow the previous example. Therefore, \(\vecs r_u \times \vecs r_v\) is not zero for any choice of \(u\) and \(v\) in the parameter domain, and the parameterization is smooth. Integration is a way to sum up parts to find the whole. Because of the half-twist in the strip, the surface has no outer side or inner side. The Divergence Theorem can be also written in coordinate form as. Calculate the Surface Area using the calculator. Not strictly required, but useful for intuition and analogy: (This is analogous to how computing line integrals is basically the same as computing arc length integrals, except that you throw a function inside the integral itself. Break the integral into three separate surface integrals. The surface in Figure \(\PageIndex{8a}\) can be parameterized by, \[\vecs r(u,v) = \langle (2 + \cos v) \cos u, \, (2 + \cos v) \sin u, \, \sin v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v < 2\pi \nonumber \], (we can use technology to verify). Use Equation \ref{scalar surface integrals}. You can also check your answers! The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). The surface integral is then. For example, the graph of paraboloid \(2y = x^2 + z^2\) can be parameterized by \(\vecs r(x,y) = \left\langle x, \dfrac{x^2+z^2}{2}, z \right\rangle, \, 0 \leq x < \infty, \, 0 \leq z < \infty\). Lets first start out with a sketch of the surface. You can accept it (then it's input into the calculator) or generate a new one. Figure 5.1. Analogously, we would like a notion of regularity (or smoothness) for surfaces so that a surface parameterization really does trace out a surface. \nonumber \]. To confirm this, notice that, \[\begin{align*} x^2 + y^2 &= (u \, \cos v)^2 + (u \, \sin v)^2 \\[4pt] &= u^2 \cos^2 v + u^2 sin^2 v \\[4pt] &= u^2 \\[4pt] &=z\end{align*}\]. The dimensions are 11.8 cm by 23.7 cm. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. &= \iint_D \left(\vecs F (\vecs r (u,v)) \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \right) || \vecs t_u \times \vecs t_v || \,dA \\[4pt] To embed this widget in a post on your WordPress blog, copy and paste the shortcode below into the HTML source: To add a widget to a MediaWiki site, the wiki must have the. Following are the steps required to use the, The first step is to enter the given function in the space given in front of the title. Use Equation \ref{equation1} to find the area of the surface of revolution obtained by rotating curve \(y = \sin x, \, 0 \leq x \leq \pi\) about the \(x\)-axis. The tangent vectors are \(\vecs t_u = \langle 1,-1,1\rangle\) and \(\vecs t_v = \langle 0,2v,1\rangle\). The antiderivative is computed using the Risch algorithm, which is hard to understand for humans. An oriented surface is given an upward or downward orientation or, in the case of surfaces such as a sphere or cylinder, an outward or inward orientation. You appear to be on a device with a "narrow" screen width (, \[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {x,y,g\left( {x,y} \right)} \right)\sqrt {{{\left( {\frac{{\partial g}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial g}}{{\partial y}}} \right)}^2} + 1} \,dA}}\], \[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}\], 2.4 Equations With More Than One Variable, 2.9 Equations Reducible to Quadratic in Form, 4.1 Lines, Circles and Piecewise Functions, 1.5 Trig Equations with Calculators, Part I, 1.6 Trig Equations with Calculators, Part II, 3.6 Derivatives of Exponential and Logarithm Functions, 3.7 Derivatives of Inverse Trig Functions, 4.10 L'Hospital's Rule and Indeterminate Forms, 5.3 Substitution Rule for Indefinite Integrals, 5.8 Substitution Rule for Definite Integrals, 6.3 Volumes of Solids of Revolution / Method of Rings, 6.4 Volumes of Solids of Revolution/Method of Cylinders, A.2 Proof of Various Derivative Properties, A.4 Proofs of Derivative Applications Facts, 7.9 Comparison Test for Improper Integrals, 9. Now consider the vectors that are tangent to these grid curves. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion. Let \(S\) be a smooth orientable surface with parameterization \(\vecs r(u,v)\). We can drop the absolute value bars in the sine because sine is positive in the range of \(\varphi \) that we are working with. Show that the surface area of cylinder \(x^2 + y^2 = r^2, \, 0 \leq z \leq h\) is \(2\pi rh\). 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 \(x^2 + y^2 + z^2 = 4\): \[x^2 + y^2 + 1 = 4 \Rightarrow x^2 + y^2 = 3. Maxima takes care of actually computing the integral of the mathematical function. If you like this website, then please support it by giving it a Like. S curl F d S, where S is a surface with boundary C. It helps you practice by showing you the full working (step by step integration). \nonumber \]. Describe the surface integral of a scalar-valued function over a parametric surface. In Example \(\PageIndex{14}\), we computed the mass flux, which is the rate of mass flow per unit area. 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. It is the axis around which the curve revolves. It also calculates the surface area that will be given in square units. By Equation, \[ \begin{align*} \iint_{S_3} -k \vecs \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_1^4 \vecs \nabla T(u,v) \cdot (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt] But, these choices of \(u\) do not make the \(\mathbf{\hat{i}}\) component zero. For more about how to use the Integral Calculator, go to "Help" or take a look at the examples. \nonumber \], From the material we have already studied, we know that, \[\Delta S_{ij} \approx ||\vecs t_u (P_{ij}) \times \vecs t_v (P_{ij})|| \,\Delta u \,\Delta v. \nonumber \], \[\iint_S f(x,y,z) \,dS \approx \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij})|| \vecs t_u(P_{ij}) \times \vecs t_v(P_{ij}) ||\,\Delta u \,\Delta v. \nonumber \]. Surface integrals are a generalization of line integrals. Volume and Surface Integrals Used in Physics. Calculate the average value of ( 1 + 4 z) 3 on the surface of the paraboloid z = x 2 + y 2, x 2 + y 2 1. Direct link to Is Better Than 's post Well because surface inte, Posted 2 years ago. Surface integrals are used anytime you get the sensation of wanting to add a bunch of values associated with points on a surface. \nonumber \], \[ \begin{align*} \iint_S \vecs F \cdot dS &= \int_0^4 \int_0^3 F (\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v) \, du \,dv \\[4pt] &= \int_0^4 \int_0^3 \langle u - v^2, \, u, \, 0\rangle \cdot \langle -1 -2v, \, -1, \, 2v\rangle \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 [(u - v^2)(-1-2v) - u] \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 (2v^3 + v^2 - 2uv - 2u) \, du\,dv \\[4pt] &= \int_0^4 \left. Therefore, we have the following equation to calculate scalar surface integrals: \[\iint_S f(x,y,z)\,dS = \iint_D f(\vecs r(u,v)) ||\vecs t_u \times \vecs t_v||\,dA. Find the surface area of the surface with parameterization \(\vecs r(u,v) = \langle u + v, \, u^2, \, 2v \rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 2\). Here is the parameterization for this sphere. Surface integrals are important for the same reasons that line integrals are important. It is the axis around which the curve revolves. To visualize \(S\), we visualize two families of curves that lie on \(S\). \end{align*}\]. This makes a=23.7/2=11.85 and b=11.8/2=5.9, if it were symmetrical. mass of a shell; center of mass and moments of inertia of a shell; gravitational force and pressure force; fluid flow and mass flow across a surface; electric charge distributed over a surface; electric fields (Gauss' Law . Parameterization \(\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\) is a regular parameterization if \(\vecs r_u \times \vecs r_v\) is not zero for point \((u,v)\) in the parameter domain. To parameterize this disk, we need to know its radius. In other words, the top of the cylinder will be at an angle. Did this calculator prove helpful to you? For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ In "Examples", you can see which functions are supported by the Integral Calculator and how to use them. 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. Calculate the area of a surface of revolution step by step The calculations and the answer for the integral can be seen here. Let \(S\) be a piecewise smooth surface with parameterization \(\vecs{r}(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle \) with parameter domain \(D\) and let \(f(x,y,z)\) be a function with a domain that contains \(S\). \nonumber \], Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Follow the steps of Example \(\PageIndex{15}\). We parameterized up a cylinder in the previous section. Parameterizations that do not give an actual surface? then, Weisstein, Eric W. "Surface Integral." Calculate surface integral \[\iint_S (x + y^2) \, dS, \nonumber \] where \(S\) is cylinder \(x^2 + y^2 = 4, \, 0 \leq z \leq 3\) (Figure \(\PageIndex{15}\)). ; 6.6.5 Describe the surface integral of a vector field. Therefore, \(\vecs t_x + \vecs t_y = \langle -1,-2,1 \rangle\) and \(||\vecs t_x \times \vecs t_y|| = \sqrt{6}\). If we want to find the flow rate (measured in volume per time) instead, we can use flux integral, \[\iint_S \vecs v \cdot \vecs N \, dS, \nonumber \]. I understood this even though I'm just a senior at high school and I haven't read the background material on double integrals or even Calc II. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. https://mathworld.wolfram.com/SurfaceIntegral.html.
Elopement Packages South Coast Nsw,
Josh Allen Career Red Zone Stats,
Foster Gillett Vail, Colorado,
Spooky Nook Tournaments,
Articles S
| how did suleika jaouad meet jon batiste | |||
| which of these best describes the compromise of 1877? | |||