Q 18. - Calculus

Question

Express the area of the region  between the function $f\left(x\right)=x^2$ and the $x$-axis on the interval $[-3, 3]$ as an iterated integral, integrating first with respect to $x$. Express the area of  as a sum of two iterated integrals, integrating first with respect to $y$ in each. Now evaluate your integrals.

Step-by-Step Solution

Answer

The Area of region is $18$ units.

Step 1: Given Information

We are given $y=x^2$ and $y=0$ and $[-3,3]$.

Step 2: Region of Integration

The region of integration is as below:

Step 3: Evaluating limits

First ${\Omega }_1$ is bounded by $x=-\sqrt{y}, x=-3$ and $x$ axis.

Hence, $-3\le x\le \sqrt{y}, 0\le y\le 9$

Similarly ${\Omega }_2$ is bounded by $x=\sqrt{y}, x=-3$ and $x$ axis.

Hence, $\sqrt{y}\le x\le 3, 0\le y\le 9$

Step 4: Area of Region by Iterated Integrals

It is written as:

${\iint }_{\Omega }dA={\iint }_{{\Omega }_1}dA+{\iint }_{{\Omega }_2}dA$

$\Rightarrow {\iint }_{\Omega }dA={\int }_0^{9-\sqrt{y}}{\int }_{-3}^{9}dxdy+{\int }_0^3{\int }_{\sqrt{y}}dxdy$

Step 5: Region of Integration

The region where first integration is done is

Step 6: Limits

From above graph

for ${\Omega }_1$, $0\le y\le x^2, -3\le x\le 0$

for ${\Omega }_2$, $0\le y\le x^2, 0\le x\le 3$

Step 7: Area as sum of integrals

Area is written as

${\iint }_{\Omega }dA={\iint }_{{\Omega }_4}dA+{\iint }_{{\Omega }_2}dA$

Using values ${\iint }_{\Omega }dA={\int }_{-3}^0{\int }_0^{x^2}dydx+{\int }_0^3{\int }_0^{x^2}dydx$

Area of region is given by

${\int }_{-3}^0{\int }_0^{x^2}dydx+{\int }_0^3{\int }_0^{x^2}dydx$

Solving integrals

${\iint }_{\Omega }dA={\int }_{-3}^0[y{]}_0^{x^2}dx+{\int }_0^3[y{]}_0^{x^2}dx$

$={\int }_{-3}^0{x}^{2}dx+{\int }_0^3{x}^{2}dx$

$={\left[\frac{x^3}{3}\right]}_{-3}^0+{\left[\frac{x^3}{3}\right]}_0^3$

$=18$

Area of region is $18$ units