# Integrate f(x, y, z) = x + \sqrt {y} - z^2 from (0, 0, 0) to (1, 1, 1) over the paths first C_1:...

## Question:

Integrate {eq}f(x, y, z) = x + \sqrt {y} - z^2 {/eq} from (0, 0, 0) to (1, 1, 1) over the paths first

{eq}C_1: r(t) = ti + t^2 j; 0 \leq t \leq 1 {/eq}

followed by {eq}C_2: i + j + tk, 0 \leq t \leq 1. {/eq}

## Definite Integral:

The definite integral is defined as an integration evaluation with the upper and the lower limits. Integrating the function using the curve over the interval is given by, {eq}\ \displaystyle \int f(x, y, z) ds =\int_{a}^{b} f(r(t))\left \| {r}'(t) \right \| dt {/eq}. We have to multiply the parameterized function and the norm of the derivative for integration.

## Answer and Explanation:

Let us consider the given fuction {eq}\displaystyle f(x, y, z) = x + \sqrt{y} - z^2 {/eq} over the paths first {eq}\displaystyle C_1: r(t) = t \vec{i} + t^2 \vec{j} + 0 \vec{k}; 0 \leq t \leq 1 {/eq}followed by {eq}\displaystyle C_2: 1 \vec{i} + 1 \vec{j} + t \vec{k}, 0 \leq t \leq 1 {/eq}.

Integrating the function using the first curve:

{eq}\begin{align*} \displaystyle f(x, y, z) &= x + \sqrt{y} - z^2 \\ \displaystyle C_1: r(t) &= t \vec{i} + t^2 \vec{j} + 0 \vec{k} \\ \displaystyle f(r(t)) &= t + \sqrt{t^2} - (0)^2 \\ \displaystyle f(r(t)) &=2t \\ \displaystyle {r}'(t) &=1 \vec{i} + 2t \vec{j} + 0 \vec{k} \\ \displaystyle \left \| {r}'(t) \right \| &=\sqrt{(1)^{2}+(2t)^{2}}=\sqrt{1+4t^2} \\ \displaystyle \int f(x, y, z) ds &=\int_{a}^{b} f(r(t))\left \| {r}'(t) \right \| dt \\ \displaystyle &=\int_{0}^{1} (2t)\left( \sqrt{1+4t^2} \right) dt \\ \displaystyle &=2 \int_{0}^{1} t\sqrt{1+4t^2} dt \\ \displaystyle \mathrm{Using\:u-substitution:} \\ \displaystyle u &=1+4t^2, \: du=8t dt, \: tdt=\frac{du}{8} \\ \displaystyle &=2 \int_{0}^{1} \sqrt{u}\frac{du}{8} \\ \displaystyle &= \frac{1}{4} \int_{0}^{1} \sqrt{u}du \\ \displaystyle &=\frac{1}{4}\left[ \frac{2u^{\frac{3}{2}}}{3} \right]_{0}^{1} \\ \displaystyle \mathrm{Substituting, \:}\:u &=1+4t^2 \\ \displaystyle &=\frac{1}{4}\left[ \frac{2(1+4t^2)^{\frac{3}{2}}}{3} \right]_{0}^{1} \\ \displaystyle &=\frac{1}{4}\left[ \left( \frac{2(1+4(1)^2)^{\frac{3}{2}}}{3} \right)-\left( \frac{2(1+4(0)^2)^{\frac{3}{2}}}{3} \right) \right] \\ \displaystyle \int f(x, y, z) ds &=1.697 \end{align*} {/eq}

Answer: {eq}\ \displaystyle \mathbf{\color{blue}{ \int f(x, y, z) ds =1.697 }} {/eq}.

Integrating the function using the second curve:

{eq}\begin{align*} \displaystyle f(x, y, z) &= x + \sqrt{y} - z^2 \\ \displaystyle C_2: r(t) &= 1 \vec{i} + 1 \vec{j} + t \vec{k} \\ \displaystyle f(r(t)) &= 1+\sqrt{1}-t^2 \\ \displaystyle f(r(t)) &= -t^2+2 \\ \displaystyle {r}'(t) &= 0 \vec{i} + 0 \vec{j} + 1 \vec{k} \\ \displaystyle \left \| {r}'(t) \right \| &=1 \\ \displaystyle \int f(x, y, z) ds &=\int_{a}^{b} f(r(t))\left \| {r}'(t) \right \| dt \\ \displaystyle &=\int_{0}^{1} (-t^2+2)(1) dt \\ \displaystyle &=\left[ -\frac{t^3}{3}+2t \right]_{0}^{1} \\ \displaystyle &=\left( -\frac{(1)^3}{3}+2(1) \right)-\left( -\frac{(0)^3}{3}+2(0) \right) \\ \displaystyle &=\frac{5}{3} \\ \displaystyle \int f(x, y, z) ds &=1.667 \end{align*} {/eq}

Answer: {eq}\ \displaystyle \mathbf{\color{blue}{ \int f(x, y, z) ds =1.667 }} {/eq}.

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13