$\int_R xy^2ds$ where $R$ is the upper half of the circle $x^2+y^2=25$.

Question

Is my procedure correct for the following problem?

Calculate

$$\int_R xy^2ds$$

where $R$ is the upper half of the circle $x^2+y^2=25$.

What I did was parametrize the circle which gives $g(t) = (5\cos(t), 5\sin(t))$ and $0 \leq t \leq \pi $. This ends in the integral

$$\int_0^{\pi} (125 \sin^2(t)\cos(t))(5)dt=0$$

Is it correct that it's $0$ or did I make an incorrect parametrization? Thanks.

Answer 1

I am going to assume that you are interested in the integral of $f(x,y) = xy^{2}$ over the region $R$: \begin{align*} I = \int_{R}xy^{2}\mathrm{d}y\mathrm{d}x = \int_{-5}^{5}\int_{0}^{\sqrt{25-x^{2}}}xy^{2}\mathrm{d}y\mathrm{d}x \end{align*}

If we make the change of variables $x = r\cos(\theta)$ and $y = r\sin(\theta)$, one gets that \begin{align*} I = \int_{0}^{5}\int_{0}^{\pi}r^{4}\cos(\theta)\sin^{2}(\theta)\mathrm{d}\theta\mathrm{d}r & = \int_{0}^{5}r^{4}\mathrm{d}r\int_{0}^{\pi}\cos(\theta)\sin^{2}(\theta)\mathrm{d}\theta = 625\times 0 = 0 \end{align*}

EDIT

If you are interested in the line integral, here it is: \begin{align*} \int_{\gamma}f(x,y)\mathrm{d}s & = \int_{0}^{\pi}f(\gamma(t))\|\gamma'(t)\|\mathrm{d}t\\\\ & = \int_{0}^{\pi}5\cos(t)\times25\sin^{2}(t)\times\sqrt{25(-\sin(t))^{2}+25\cos^{2}(t)}\mathrm{d}t\\\\ & = 625\times\int_{0}^{\pi}\cos(t)\sin^{2}(t)\mathrm{d}t = 0 \end{align*}

Answer 2

It's unclear whether you want the surface integral over the region enclosed or the line integral over the boundary of the semicircle, but you can use a symmetry argument in any case.

Here is how you would do it for the surface integral. Let us split your region into two disjoint parts $A$ and $B$, where $A = \{(x, y) ~|~ x = 0\}$ and $B = \{(x, y) ~|~ x \neq 0\}$. Note that for every point $(x, y)$ with $x \neq 0$ in region $B$, the point $(-x, y)$ also lies in the region. Since the integrand is odd in $x$ (i.e. $xy^2 = -((-x)y^2$), and hence $\int_B xy^2 ds = 0$. In addition, the integrand is zero on every point in $A$, so it follows that $\int_A xy^2 ds = 0$. Hence, $$\int_R xy^2 ds = \int_A xy^2 ds + \int_B xy^2 ds = 0$$ The symmetry argument for the line integral is similar.