Application of Double Integrals

Question

Find the volume of the solid bounded above by $z=4-y^2-\frac{1}{4}x^2$ and below by the disk $(y-1)^2+x^2 \le 1$ , $z\ge0$.

I first converted the equation $z=4-y^2-\frac{1}{4}x^2$ into polar equation, which then became $z=4-r^2\sin^2\theta-\frac{r^2\cos^2\theta}{4}$. Then I also converted the equation of the disk which then became $r=2\sin\theta$. To get the volume of the solid I evaluated the double integral $$\int^{\pi}_{0}\int^{2\sin\theta}_{0} \left(4-r^2\sin^2\theta-\frac{r^2\cos^2\theta}{4}\right) rdrd\theta.$$ The answer I got in the end was $\frac{63\pi}{16}-\frac{16}{3}$. Is my answer correct? If it is not, can someone point out what part/s I was/were wrong. Thank you in advance.

Answer 1

Using $x = r \cos \theta$ and $y=1+r \sin \theta$, you get $z = 4-(1+ r \sin \theta)^2- \frac 14 r^2 \cos^2 \theta$. In these coordinates, $$ \int_0^{2 \pi} \int_0^1 r(4-(1+ r \sin \theta)^2- \frac 14 r^2 \cos^2 \theta) dr d\theta =\frac{43 \pi}{16}. $$

Answer 2

The issue is with your last integration. Your formula is right, but you got the wong answer $$\int^{\pi}_{0}\int^{2\sin\theta}_{0} \left(4-r^2\sin^2\theta-\frac{r^2\cos^2\theta}{4}\right) rdrd\theta= \\ =\int_0^\pi\left(\frac424\sin^2\theta-\frac14(2\sin\theta)^4\sin^2\theta-\frac14(2\sin\theta)^4\frac 14\cos^2\theta)\right)d\theta\\=\int_0^\pi(8\sin^2\theta-\frac1{4}16\sin^6\theta-\frac1{16}16\sin^4\theta\cos^2\theta)d\theta\\=\frac{43\pi}{16}$$

Answer 3

There is nothing wrong with your method. It is definitely more tedious as the upper bound of the first integral will be $2 \sin \theta$. Also, you are taking the angle from the origin for a disk with center at $(0,1)$ and radius $1$, and hence $0$ to $\pi$ will be the bound for the whole disk area.

$ \displaystyle \int^{\pi}_{0}\int^{2\sin\theta}_{0} \left(4-r^2\sin^2\theta-\frac{r^2\cos^2\theta}{4}\right) rdrd\theta = \displaystyle \int^{\pi}_{0}[2r^2-\frac{r^4}{4} \sin^2\theta - \frac{r^4}{16} \cos^2\theta]_0^{2\sin\theta} \, d\theta$

$\displaystyle \, = \int_0^{\pi} (8\sin^2 \theta - \sin^4 \theta \cos^2 \theta - 4 \sin^6 \theta) d \theta = \frac{43 \pi}{16}$

May be you missed to multiply by $r$ before first integration ($r dr$) and hence got a wrong answer.