# Recent questions tagged area-under-curve

A parabola $x=y^2$ with $0 \leq x \leq 1$ is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by $360^{\circ}$ around x-axis is $\dfrac{\pi}{4} \\$ $\dfrac{\pi}{2} \\$ ${\pi} \\$ $2 \pi$
The area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x-y$ plane is $1/6$ $1/4$ $1/3$ $1/2$
A parametric curve defined by $x= \cos \left ( \dfrac{\Pi u}{2} \right ), y= \sin \left ( \dfrac{\Pi u}{2} \right )$ in the range $0 \leq u \leq 1$ is rotated about the $X$-axis by $360$ degrees. Area of the surface generated is $\dfrac{\Pi }{2} \\$ $\pi \\$ $2 \pi \\$ $4 \pi$
The surface integral $\displaystyle{} \int \int_{s}^{ }\dfrac{1}{\pi }(9xi-3yj)\cdot nds$ over the sphere given by $x^2+y^2+z^2=9$ is
The integral $\oint_{c}^{ } (ydx-xdy)$ is evaluated along the circle $x^2+y^2=\frac{1}{4}$ traversed in counter clockwise direction. The integral is equal to $0$ $\frac{-\pi }{4}$ $\frac{-\pi }{2}$ $\frac{\pi }{4}$
The following surface integral is to be evaluated over a sphere for the given steady velocity vector field $F$ = $xi$ + $yj$+ $zk$ defined with respect to a Cartesian coordinate system having $i$, $j$ and $k$ as unit base vectors. $\iint_{s}^{ }\frac{1}{4}(F.n)dA$ ... $\pi$ $2\pi$ $3\pi/4$ $4\pi$