Evaluate the surface integral {eq}\int\int_SzdS{/eq}, where S is the plane x+y+2z=5 in the first octant
Question:
Evaluate the surface integral {eq}\int\int_SzdS{/eq}, where S is the plane x+y+2z=5 in the first octant
Line integral concept:
You have to evaluate the surface integral where S is the given plane in the first octant. It have a surface S defined by the intersection of the plane ax + by + cz = d with the first octant, where a,b,c,d are positive. The intersection of the plane with the xy plane is defined by z=0. Solve this equation for y and it gives the upper bound on y. Next set y =0 and solve for x .This gives the upper bound for x. Find the first derivative and substitute the values in below formula
{eq}\int {\int_S {f\left( {x,y,z} \right)} } dS = \int_c^d {\int_a^b {f\left( {x,y} \right)\sqrt {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2} + {{\left( {z'} \right)}^2}} dydx} }. {/eq}
Answer and Explanation: 1
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View this answerThe line segment has an equation
{eq}x+y+2z=5. {/eq}
Find the first derivative and you have
{eq}x' = 1\\ y' = 1\\ z' = 2 {/eq}
Now, evaluate...
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Line integrals are any integral of a function that can be defined along a given curve in a three-dimensional space. Learn the process of line integration and how they can be used to map paths using parametrizations.
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