(b) Set up and evaluate $$ \iint_{D} e^{y^{3}} d A $$ where $$ D $$ is the region bounded by $$ y=\sqrt{x}, y=2 $$ and $$ x=0 $$

Question

UpStudy AI · Accepted Answer

We first describe the region $$D$$. The region is bounded by
- $$y=\sqrt{x}$$ (which can be rewritten as $$x=y^2$$),
- $$x=0$$, and
- $$y=2$$.

For a fixed $$y$$ between $$0$$ and $$2$$, the variable $$x$$ runs from $$0$$ to $$y^2$$. Hence, we set up the integral as

$$
\int_{y=0}^{2}\int_{x=0}^{y^2} e^{y^3}\,dx\,dy.
$$

Since $$e^{y^3}$$ does not depend on $$x$$, the inner integral is

$$
\int_{x=0}^{y^2} e^{y^3}\,dx = e^{y^3} \left[x\right]_{0}^{y^2} = y^2 e^{y^3}.
$$

Thus, the double integral becomes

$$
\int_{0}^{2} y^2 e^{y^3}\,dy.
$$

To evaluate this, use the substitution $$u=y^3$$. Then

$$
\frac{du}{dy}=3y^2 \quad \Longrightarrow \quad y^2\,dy=\frac{du}{3}.
$$

The limits change as follows:
- When $$y=0$$, $$u=0^3=0$$.
- When $$y=2$$, $$u=2^3=8$$.

Substituting, the integral becomes

$$
\int_{0}^{2} y^2 e^{y^3}\,dy = \frac{1}{3}\int_{u=0}^{8} e^u\,du.
$$

Evaluating the integral of $$e^u$$,

$$
\frac{1}{3}\int_{0}^{8} e^u\,du = \frac{1}{3}\left(e^u\right)\Big|_{0}^{8} = \frac{1}{3} \left(e^8 - 1\right).
$$

Hence, the value of the double integral is

$$
\frac{e^8-1}{3}.
$$
The value of the double integral $$ \iint_{D} e^{y^{3}} \, dA $$ is $$ \frac{e^{8} - 1}{3} $$.

(b) Set up and evaluate where is the region bounded by and