Use geometry to compute the given integral.
E)none of these

Question

Use geometry to compute the given integral. 
E)none of these

Quizplus · Accepted Answer

The given integral can be written as:
$$\int_{-1}^1\int_{0}^{\sqrt{1-x^2}}\int_{0}^{x^2+y^2}dzdydx$$
The limits of $z$ are from $0$ to $x^2+y^2$, which implies that the region of integration lies below the surface $z=x^2+y^2$. This is a circular paraboloid with vertex at the origin and axis along the $z$-axis.

The limits of $y$ are from $0$ to $\sqrt{1-x^2}$, which implies that the region of integration lies between the $xy$-plane and the upper half of the unit circle centered at the origin which lies in the $xy$-plane.

The limits of $x$ are from $-1$ to $1$, which implies that the region of integration lies between the planes $x=-1$ and $x=1$.

Therefore, we can evaluate the integral by computing the triple integral in the order $dzdydx$ as follows:

\begin{align*}
\int_{-1}^1\int_{0}^{\sqrt{1-x^2}}\int_{0}^{x^2+y^2}dzdydx &= \int_{-1}^1\int_{0}^{\sqrt{1-x^2}}\left[x^2+y^2ight]dydx\
&=\int_{-1}^1\left[x^2y+\frac{y^3}{3}ight]_{0}^{\sqrt{1-x^2}}dx\
&=\int_{-1}^1\left[x^2\sqrt{1-x^2}+\frac{1}{3}(1-x^2)^\frac{3}{2}ight]dx\
&=\frac{\pi}{4}.
\end{align*}

Use Geometry to Compute the Given Integral