Using double integral to find the volume of region

Question

Question

Find the volume of solid which is enclosed by cylinders $$x^2+y^2=2ay$$ And $$z^2=2ay$$ $x \geq 0$ $z \geq 0$

My attempt

Using double integral As $ 0\leq y \leq 2a$ and $0 \leq x \leq \sqrt{2ay-y^2}$ $$I=\int_0^{2a} \int_0^{\sqrt{2ay-y^2}} \sqrt{2ay} dy dx$$ Forming it after integrating with respect to $x$ as

$$I =\sqrt{2a} \int_0^{2a} \sqrt{y+a-a} \sqrt{a^2-(y-a)^2}$$ Now let $y-a = a \sin{\theta}$ Therefore $ dy = a \cos{\theta} d\theta $ So the integral $I$ becomes $$I=\sqrt{2} a^3 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1+ \sin{\theta}} {cos^{2}{\theta}} d{\theta}$$ From here i have no idea either i go for half angle or any other substitution.

Suggestions are appreciated

Thank you

Answer 1

You have the integrands backwards. It should be $dx\,dy,$ not $dy\,dx.$ Fixing that, you get the internal integral:

$$ \int_{0}^{\sqrt{2ay-y^2}} \sqrt{2ay} dx=\sqrt{2ay-y^2}\sqrt{2ay}=2ay\sqrt{1-\frac{y}{2a}}$$

because the integrand is constant for all $x$ in the given range.

Then you want:

$$I=2a\int_0^{2a}y\sqrt{1-\frac{y}{2a}}\,dy$$

Letting $u=1-\frac{y}{2a}$ then $y=2a(1-u)$ and $dy=-2a\,du,$ so you get:

$$I=-8a^3\int_{1}^{0} (1-u)\sqrt{u}\,du=8a^3\int_{0}^{1} (1-u)\sqrt{u}\,du$$

And $$\int (1-u)\sqrt{u}\,du = \frac{2}{3}u^{3/2}-\frac{2}{5}u^{5/2}$$

so we get:

$$I=8a^3\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{64a^3}{15}.$$

Answer 2

From your second last line, you can put everything under a single radical, to get that we’re integrating $\int y\sqrt{2a-y}dy$. From here you can let $u=2a-y$ to finish the problem.