The value of ${\int }_0^{\pi /2}$${\int }_0^{cos\theta }$r sin θ dr dθ is

The correct option is 1/6.

Step-by-Step Explanation:

As shown in the provided image, we are given the double integral:
$$I={\int }_0^{\pi /2}{\int }_0^{\cos \theta }r\sin \theta drd\theta$$
The order of the differentials $drd\theta$ indicates the priority of integration. The limits of the inner integral for $r$ range from $0$ to $\cos \theta$, and the limits of the outer integral for $\theta$ range from $0$ to $\pi /2$.

Step 1: Evaluate the inner integral with respect to $r$
Since the integration is with respect to $r$, the term $\sin \theta$ is treated as a constant. We can factor it out of the inner integral:
$${\int }_0^{\cos \theta }r\sin \theta dr=\sin \theta {\int }_0^{\cos \theta }rdr$$
Using the power rule for integration, $\int rdr=\frac{r^2}{2}$:
$$\sin \theta {\left[\frac{r^2}{2}\right]}_0^{\cos \theta }=\sin \theta (\frac{{\cos }^2\theta }{2}-0)=\frac{1}{2}{\cos }^2\theta \sin \theta$$

Step 2: Evaluate the outer integral with respect to $\theta$
Now we substitute this result back into the main integral:
$$I=\frac{1}{2}{\int }_0^{\pi /2}{\cos }^2\theta \sin \theta d\theta$$
To evaluate this integral, we use the method of substitution. Let:
$$u=\cos \theta$$
Differentiating both sides gives:
$$du=-\sin \theta d\theta \Rightarrow \sin \theta d\theta =-du$$
We must also change the limits of integration accordingly:
• When $\theta =0$, $u=\cos \left(0\right)=1$.
• When $\theta =\pi /2$, $u=\cos (\pi /2)=0$.
Substituting these values in, we get:
$$I=\frac{1}{2}{\int }_1^0{u}^2(-du)$$
Using the property of integrals to flip the limits of integration by changing the sign of the integrand, we obtain:
$$I=\frac{1}{2}{\int }_0^1{u}^{2}du$$
Now evaluate the definite integral:
$$I=\frac{1}{2}{\left[\frac{u^3}{3}\right]}_0^1=\frac{1}{2}(\frac{1^3}{3}-0)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$$
Thus, the value of the double integral is indeed 1/6.