# evaluate the integrals \int \int y\sqrt{-20+12y-y^2} dy

## Question:

evaluate the integrals \int

{eq}\int y\sqrt{-20+12y-y^2} dy{/eq}

## Integration.

There are some relation between differentiation and integration and this is decided by fundamental theorem of integration.

It defines the limit of Riemann's theorem.

The formula is:

{eq}\displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ \displaystyle\int \sqrt {a^2-x^2}dx=\dfrac{x\sqrt {a^2-x^2}}{2}+\dfrac{a^2}{2}sin^{-1}\left ( \dfrac{x}{a} \right )+c\\\\ {/eq}

We have to calculate the given integrals:

{eq}\displaystyle\int y\sqrt {-20+12y-y^2}dy\\\\ {/eq}

We may write as:

{eq}-20+12y-y^2=-(y-6)^2+16\\\\ =\displaystyle\int y\sqrt {-(y-6)^2+16}dy\\\\ {/eq}

Let:

{eq}y-6=z\\\\ dy=dz\\\\ y=z+6\\\\ {/eq}

Hence:

{eq}=\displaystyle\int (z+6)\sqrt {-z^2+16}dz\\\\ \displaystyle\int z\sqrt {-z^2+16}dz+\displaystyle\int 6\sqrt {-z^2+16}dz\,\,\,\,\,eqn(1)\\\\ {/eq}

Hence integrating separately:

{eq}\displaystyle\int z\sqrt {-z^2+16}dz\\\\ {/eq}

Let:

{eq}-z^2+16=t\\\\ -2z\ dz=dt\\\\ z\ dz=\dfrac{-dt}{2}\\\\ {/eq}

{eq}=\displaystyle\int \sqrt {t}\cdot \left ( \dfrac{-dt}{2} \right )\\\\ =\dfrac{-1}{2}\cdot \dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}\\\\ =\dfrac{-1}{2}\cdot \dfrac{2}{3}(-z^2+16)^{\frac{3}{2}}\\\\ =\dfrac{-1}{3}\left ( -z^2+16 \right )^{\frac{3}{2}}\\\\ \displaystyle\int 6\sqrt {-z^2+16}dz\\\\ {/eq}

Therefore integrating the second part:

{eq}6\displaystyle\int \sqrt {\left ( 4 \right )^2-\left ( z \right )^2}dz\\\\ \displaystyle\int \sqrt {a^2-x^2}dx=\dfrac{x\sqrt {a^2-x^2}}{2}+\dfrac{a^2}{2}sin^{-1}\left ( \dfrac{x}{a} \right )+c\\\\ =6\cdot \dfrac{z\sqrt {4^2-z^2}}{2}+6\cdot \dfrac{16}{2}sin^{-1}\left ( \dfrac{z}{4} \right )+c\\\\ =3(y-6)\sqrt {16-(y-6)^2}+48sin^{-1}\left ( \dfrac{y-6}{4} \right )+c\\\\ {/eq}

Substituting the above values in eqn(1) we get:

{eq}=\dfrac{-1}{3}(-z^2+16)^{\frac{3}{2}}+3(y-6)\sqrt {16-(y-6)^2}+48sin^{-1}\left ( \dfrac{y-6}{4} \right )+c\\\\ {/eq}