Exam 14: Iterated Integrals and Area in the Plane

Find the centroid of the solid region bounded by the graphs of the equations. Use a computer algebra system to evaluate the triple integral. (Assume uniform density and find the center of mass.) $z = \frac { 9 } { y ^ { 2 } + 1 } , z = 0 , x = - 2 , x = 2 , y = 0 , y = 1$

Evaluate the following iterated integral. $\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{2}^{4} \rho^{2} \sin \phi d \rho d \phi d \theta$

Find the mass of the lamina described by the inequalities $0 \leq x \leq 4 \text { and } 0 \leq y \leq 8 \text {, }$ given that its density is $\rho ( x , y ) = 2 x y$ . (Hint: Some of the integrals are simpler in polar coordinates.)

Evaluate the following integral. $\int _ { 4 x } ^ { x ^ { 6 } } \frac { - 8 y } { x } d y$

Consider the region R in the xy-plane bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{4}$ =1 transformation $x = 4 u$ and $y = 2 v$ . Find the area of the ellipse. Round your answer to two decimal places.

Evaluate the following iterated integral. $\int_{0}^{\pi /15}\int_{0}^{\pi/15}\int_{0}^{cos\theta}\rho^{2} \sin \phi \cos \phi d \rho d \theta d \phi$

Use cylindrical coordinates to find the volume of the solid inside both $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 256$ and $( x - 8 ) ^ { 2 } + y ^ { 2 } = 64$ .

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral below over the region R. $\iint _ { R } \frac { y } { x ^ { 2 } + y ^ { 2 } } d A$ $R$ : triangle bounded by $y = 2 x , y = 6 x$ , and $x = 5$

Evaluate the double integral below. $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 3 } 2 r ^ { 5 } \sin \theta d r d \theta$

$\text { Find the area of the portion of the surface } z = 8 x + 4 y \text { that lies above the triangular }$ region with vertices $( 0,0 ) , ( 2,0 )$ , and $( 0,2 )$ .

Evaluate the improper iterated integral $\int_{0} ^ {\infty}$$\int_{0}^{\infty} x y e^{-\left(x^{2}+y^{2}\right)} d x d y$

Use an iterated integral to find the area of the region bounded by the graphs of the equations $y = 22 - x ^ { 2 } \text { and } y = 2 x + 7$

The area of a region R is given by the iterated integral $\int_{0}^{8} \int_{\frac{x}{8}}^{1} d y d x$ Switch the order of integration and show that both orders yield the same area. What is this area?

Use a double integral to find the area of the shaded region as shown in the figure below.

$\text { Evaluate } \int _ { 0 } ^ { 8 } \int _ { \frac { 1 } { 5 } y } ^ { \sqrt { y } } x ^ { 2 } y ^ { 2 } d x d y \text {. Round your answer to two decimal places. }$

Set up and evaluate a double integral to find the volume of the solid bounded by the graphs of the equations given below. $z = x y ^ { 2 } , z > 0 , x > 0,5 x < y < 2$

Sketch the region R of integration and then switch the order of integration for the following integral. $\int _ { 0 } ^ { 7 } \int _ { 0 } ^ { \sqrt { 49 - x ^ { 2 } } } f ( x , y ) d y d x$

Find the area of the surface given by $f ( x , y ) = 5 - x ^ { 2 }$ $R$ : rectangle with vertices $( 0,0 ) , ( 5,0 ) , ( 5,5 ) , ( 0,5 )$

Evaluate the following iterated integral. $\int _ { 2 } ^ { 3 } \int _ { y } ^ { 7 y } \left( 7 + 3 x ^ { 2 } + 3 y ^ { 2 } \right) d x d y$

$\text { Evaluate the iterated integral } \int _ { 0 } ^ { \pi / 2 } \int _ { 0 } ^ { 8 } \int _ { 0 } ^ {e^ {- y ^ { 2 }} } r d z d r d \theta \text {. }$