I must find the volume of the solid that is bounded by:
$z = 0$
$z = 4 – y^2$
$x^2 + y^2 = 2x$
So far, I understand that the equation of $z = 0$ is $xy$ plane. The equation $z = 4 – y^2$ passes through the points $y = -2, y = 2$ when $z = 0$. So that gives us the solid that is under the 2nd equation, and above the 1st. So I have $-2 \leq y \leq 2$
The 3rd equation is a cylinder. Here's where I get confused.
How am I supposed to find the bounds of x? When I do some algebra, I find that $ 1 – \sqrt(1 – y^2) \leq x \leq \sqrt(1-y^2) + 1$
So I evaluate the double integral:
$\int_{-2}^{2}\int_{1 – \sqrt(1 – y^2)}^{\sqrt(1-y^2) + 1} 4 – y^2 dxdy$
This looks ugly, and I know it's wrong because when evaluated it gives a complex number. My teacher said that I should evaluate this integral using polar coordinates. Why is this? And why isn't my method working?
In general I'm having trouble solving these kinds of problems. Can anyone provide me with some general guidelines as to how to solve them?
Best Answer
As noted in the OP the equation in $x,y$ can be rewritten as $$(x-1)^2+y^2=1^2\ ,$$ it is a circle centered in $(1,0)$ with radius one in the $xOy$ plane. So we have to compute the integral from $4-y^2$ on this circle. The best way to do it is to use polar coordinates (centered in $(1,0)$): $$ \begin{aligned} x &= 1+r\cos t\ ,\\ y &= 0+r\sin t\ ,\\ dx\; dy &\cong dy\wedge dy \\ &=(\cos t\; dr-r\sin t\; dt)\wedge (\sin t\; dr+r\cos t\; dt) \\ &=r\; dr\wedge dt \\ &=r\; dr\; dt\ , \end{aligned} $$ so we have to compute $$ \begin{aligned} \iiint_{\text{given domain}}dx\; dy\;dz &= \int_0^1 r\;dr \int_0^{2\pi} (4-r^2\sin^2 t)\; dt \\ &= \int_0^1 r\;dr \;(8\pi -\pi r^2) \\ &=\frac {15}4\pi\ . \end{aligned} $$