Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D xy^2\,dx\,dy, $$ where the region of integration is bounded by the parabola $y^2=2x$ and the line $x=\frac12$.
Explanation of Method
The parabola $y^2=2x$ opens rightward; the vertical line $x=\frac12$ cuts it at $y=\pm1$. Integrating with $x$ as the inner variable (from the parabola out to the line) and $y$ as the outer variable makes the inner integral a simple polynomial in $x$.
- Find the intersection points of the parabola and the line.
- Set up the integral with $x$ inner, $y$ outer.
- Integrate with respect to $x$, then $y$, using the evenness of the integrand in $y$.
Step 1 · Identify the region
$D$ is bounded on the left by the parabola $x=\frac{y^2}{2}$ and on the right by the line $x=\frac12$.
Step 2 · Find the intersection points
At $x=\frac12$: $y^2=2\cdot\frac12=1\Rightarrow y=\pm1$. Intersection points: $\left(\frac12,1\right)$ and $\left(\frac12,-1\right)$.
Step 3 · Set up the integral
$$ I = \int_{-1}^{1}\int_{y^2/2}^{1/2} xy^2\,dx\,dy. $$
Step 4 · Integrate with respect to $x$
$$ \int_{y^2/2}^{1/2} xy^2\,dx = y^2\left[\frac{x^2}{2}\right]_{y^2/2}^{1/2} = y^2\left(\frac18-\frac{y^4}{8}\right) = \frac{y^2(1-y^4)}{8} = \frac{y^2-y^6}{8}. $$
Step 5 · Integrate with respect to $y$
Since $y^2$ and $y^6$ are even functions:
$$ I = \frac18\int_{-1}^{1}(y^2-y^6)\,dy = \frac14\int_0^1(y^2-y^6)\,dy = \frac14\left[\frac{y^3}{3}-\frac{y^7}{7}\right]_0^1 = \frac14\left(\frac13-\frac17\right) = \frac14\cdot\frac{4}{21} = \frac{1}{21}. $$
Additional Notes
- Choosing $y$ as the outer variable avoids splitting the region into an upper and lower branch of the parabola, since $x=\frac{y^2}{2}$ is already a single-valued function of $y$.
Result
The solution to the integral is: